3 Marks Question - Physics STD 12 Science Questions - Vidyadip

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Question 13 Marks

Obtain the expression for the cyclotron frequency.
A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer.

Answer

$\frac{mv^2}{r}=qvB$

$r=\frac{mv}{qB}$

Frequency of revolution $(v)=\frac{1}{Time\text{ }period}=\frac{v}{2\pi r}$

$v=\frac{qB}{2\pi m}$

No The mass of the two particles, i.e deuteron and proton, is different. Since (cyclotron) frequency depends inversely on the mass, they cannot be accelerated by the same oscillator frequency.

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Question 23 Marks

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer

Length of a side of the square coil, l = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, $\theta=30^\circ$
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
$\text{T}=\text{n}\text{BIA}\sin\theta$
Where,
A = Area of the square coil
$\Rightarrow\text{l}\times\text{l}=0.1\times0.1=0.01\ \text{m}^{2}$
$\therefore\text{T}=20\times0.8\times12\times0.01\times\sin\theta$
$=0.96\ \text{N m}$
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.

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Question 33 Marks

The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Answer

Current in both wires, I = 300 A
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, l= 70 cm = 0.7 m
Force between the two wires is given by the relation,
$\text{F}=\frac{\mu_{0}\text{I}^{2}}{2\pi\text{r}}$
Where,
$\mu_{0}=\text{Permeability of free space}=4\pi\times10^{-7}\text{T m A}^{-1}$
$\therefore\text{F}=\frac{4\pi\times10^{-7}\times(300)^{2}}{2\pi\times0.015}$
$=1.2\ \text{N/m}$
Since the direction of the current in the wires is opposite, a repulsive force exists between them.

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Question 43 Marks

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer

Number of turns on the circular coil, n = 100
Radius of each turn, r = 8.0 cm = 0.08 m
Current flowing in the coil, I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,
$|\text{B}|=\frac{\mu_{0}}{4\pi}\frac{2\pi\text{nl}}{\text{r}}$
Where,
$\mu_{0}$ = Permeability of free space
$=4\sqcap\times10^{-7}\text{T m A}^{-1}$
$|\text{B}|=\frac{4\pi\times10^{-7}}{4\pi}\times\frac{2\pi\times100\times0.4}{0.08}$
$=3.14\times10^{-4}\text{T}$
Hence, the magnitude of the magnetic field is $3.14 \times 10^{-4} T.$

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Question 53 Marks

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer

Magnetic field strength, B = 6.5 \times 10^{-4} TCharge of the electron, e $= 1.6 \times 10^{-19}\ C$
Mass of the electron, me $= 9.1 \times 10^{-31}\ kg$
Velocity of the electron, $v = 4.8 \times 10^6\ m/s$
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = v
Angular frequency of the electron $=\omega=2\pi\text{v}$
Velocity of the electron is related to the angular frequency as:
$\text{v}=\text{r}\omega$
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
$\text{ev B}=\frac{\text{mv}^{2}}{\text{r}}$
$\text{eB}=\frac{\text{m}}{\text{r}}(\text{r}\omega)=\frac{\text{m}}{\text{r}}(\text{r}2\pi\text{v})$
$\text{v}=\frac{\text{B}e}{2\pi\text{m}}$
This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:
$\text{v}=\frac{6.5\times10^{-4}\times1.6\times10^{-19}}{2\times3.14\times9.1\times10^{-31}}$
$=18.2\times10^{6}\text{Hz}$
$=18\ \text{MHz}$
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of theelectron.

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Question 63 Marks

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Answer

Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
$\therefore$ Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the maqnetic field at that point is given by the relation,$\text{B}=\frac{\mu_{0}2\text{l}}{4\pi\text{r}}$
Where,
$\text{B}=\frac{\mu_{0}}{4\pi}\frac{2\text{l}}{\text{r}}$
Where,
$\mu_{0}$ = Permeability of free space $=4\sqcap\times10^{-7}\text{T m A}^{-1}$
$\text{B}=\frac{4\pi\times10^{-7}\times2\times50}{4\pi\times2.5}$
$=4\times10^{-6}\text{T}$
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell's right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

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Question 73 Marks

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer

Number of turns on the circular coil, n = 30

Radius of the coil, r = 8.0 cm = 0.08 m

Area of the coil $=\pi\text{r}^{2}=\pi(0.08)^{2}=0.0201$

Current flowing in the coil, I = 6.0 A

Magnetic field strength, B = 1 T

Angle between the field lines and normal with the coil surface,
θ = 60°

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

$\text{T}=\text{nIBA}\sin\theta...(\text{i})$

= 30 × 6 × 1 × 0.0201 × sin60°

= 3.133 N m

It can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

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Question 83 Marks

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the

Total torque on the coil,
Total force on the coil,
Average force on each electron in the coil due to the magnetic field.

$($The coil is made of copper wire of cross-sectional area $10^{–5} m^{2}$ , and the free electron density in copper is given to be about $10^{29} m^{–3}).$

Answer

Number of turns on the circular coil, n = 20 Radius of the coil, r = 10 cm = 0.1 m Magnetic field strength, B = 0.10 T Current in the coil, I= 5.0 A

The total torque on the coil is zero because the field is uniform.
The total force on the coil is zero because the field is uniform.
Cross-sectional area of copper coil, $A= 10^{-5} m^2.$

Number of free electrons per cubic meter in copper, $N = 10^{29}/m^3.$ charqe on the electron, $e = 1.6 \times 10^{-19} c$ Magnetic force, $F = Bev_d$ Where,$V_d =$ Drift velocity of electrons
$=\frac{\text{I}}{\text{NeA}}$ $\therefore\text{F}=\frac{\text{BeI}}{\text{NeA}}$ $=\frac{0.10\times5.0}{10^{29}\times10^{-5}}=5\times10^{-25}\text{N}$ Hence, the average force on each electron is $5\times10^{-25}\text{N}$.

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Question 93 Marks

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer

Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m
Hence, magnetic field at that point is given by the relation,
$\text{B}=\frac{\mu_{0}2\text{l}}{4\pi\text{r}}$
Where,
$\mu_{0}$ = Permeability of free space $=4\sqcap\times10^{-7}\text{T m A}^{-1}$
$\text{B}=\frac{4\pi\times10^{-7}\times2\times90}{4\pi\times1.5}$
$=1.2\times10^{-5}\text{T}$
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell's right hand thumb rule, the direction of the magnetic field is towards the South.

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Question 103 Marks

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer

Current flowing in wire A, $I_A = 8.0 A$
Current flowing in wire B, $1_B = 5.0 A$
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l = 10 cm = 0.1 m
Force exerted on length I due to the magnetic field is given as:
$\text{B}=\frac{\mu_{0}2\text{I}_{\text{A}}\text{I}_{\text{B}}\text{l}}{4\pi\text{r}}$
Where,
$\mu_{0}=$ Permeability of free space $= 4n \times 10^{-7}\ T\ m\ A^{-1}$
$\text{B}=\frac{4\pi\times10^{-7}\times2\times8\times5\times0.1}{4\pi\times0.04}$
$=2\times10^{-5}\ \text{N}$
The magnitude of force is $ 2 \times 10^{-5} N$. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

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Question 113 Marks

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Answer

Resistance of the qalvanorneter coil, G = 12 Ω
Current for which there is full scale deflection, $I_{g }=3\ mA = 3 \times 10^{-3} A$
Range of the voltmeter is 0, which needs to be converted to 18 V.
$\therefore\text{V}=18\text{V}$
Let a resistor of resistance R be connected in series with the galvanorneter to convert it into a voltmeter. This resistance is given as:
$\text{R}=\frac{\text{V}}{\text{I}_{\text{g}}}-\text{G}$
$=\frac{18}{3\times10^{-3}}-12=6000-12-5988\Omega$
Hence, a resistor of resistance 5988 Ω is to be connected in series with the galvanometer.

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Question 123 Marks

A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field,

Outside the toroid,
Inside the core of the toroid and,
In the empty space surrounded by the toroid.

Answer

Inner radius of the toroid, $r_1 = 25 cm = 0.25\ m$
Outer radius of the toroid,$ r_2 = 26 cm = 0.26\ m$
Number of turns on the coil, N = 3500
Current in the coil, I = 11 A

Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.
Magnetic field inside the core of a toroid is given by the relation.

$\text{B}=\frac{\mu_{0}\text{NI}}{l}$

Where,

$\mu_{0}=\text{Permeability of free space}=4\pi\times10^{-7}\text{T mA}^{-1}$

l = length of toroid

$2\pi\Big[\frac{\text{r}_{1}+\text{r}_{2}}{2}\Big]$

$\pi(0.25+0.26)$

$\therefore\text{B}=\frac{4\pi\times10^{-7}\times3500\times11}{0.51\pi}$

$≈ 3.0 \times 10^{-2}\ T$

Magnetic field in the empty space surrounded by the toroid is zero.

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Question 133 Marks

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Answer

Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, $\theta$ = 30°.
Magnetic force per unit length on the wire is given as:
$\text{f}=\text{BI}\ \sin\theta$
$=0.15\times8\times1\times\sin30^\circ$
$=0.6\text{N m}^{-1}$
Hence, the magnetic force per unit length on the wire is $0.6 N\ m^{-1}.$

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Question 143 Marks

Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1 = 10 Ω, N_1 = 30,$
$A_1 = 3.6 \times 10^{–3} m^2, B_1 = 0.25 T$
$R_2 = 14 Ω, N_2 = 42,$
$A_2 = 1.8 \times 10^{–3} m^2, B_2 = 0.50 T$
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_2$ and $M_1.$

Answer

For moving coil meter M:
Resistance, $\mathrm{R}_1=10 \Omega$
Number of turns, $\mathrm{N}_1=30$
Area of cross-section, $\mathrm{A}_1=3.6 \times 10^{-3} \mathrm{~m}^2$
Magnetic field strength, $\mathrm{B}_1=0.25 \mathrm{~T}$
Spring constant $\mathrm{K}_1=\mathrm{K}$
For moving coil meter $\mathrm{M}_2$ :
Resistance, $\mathrm{R}_2=14 \Omega$
Number of turns, $\mathrm{N}_2=42$
Area of cross-section, $\mathrm{A}_2=1.8 \times 10^{-3} \mathrm{~m}^2$
Magnetic field strength, $\mathrm{B}_2=0.50 \mathrm{~T}$
Spring constant, $\mathrm{K}_2=\mathrm{K}$

Current sensitivity of $M_1$ is given as:

$\text{l}_{s1}=\frac{\text{N}_{1}\text{B}_{1}\text{A}_{1}}{\text{K}_{1}}$

And, current sensitivity of $M_2$ is given as:

$\text{l}_{s2}=\frac{\text{N}_{2}\text{B}_{2}\text{A}_{2}}{\text{K}_{2}}$

$\therefore\text{Ratio}\frac{\text{l}_{\text{s}2}}{\text{l}_{\text{s}1}}=\frac{\text{N}_{2}\text{B}_{2}\text{A}_{2}\text{K}_{1}}{\text{K}_{2}\text{N}_{1}\text{B}_{1}\text{A}_{1}}$

$=\frac{42\times0.5\times1.8\times10^{-3}\times\text{K}}{\text{K}\times30\times0.25\times3.6\times10^{-3}}=1.4$

Voltage sensitivity for $M_2$ is given as:

$\text{V}_{s2}=\frac{\text{N}_{2}\text{B}_{2}\text{A}_{2}}{\text{K}_{2}\text{R}_{2}}$

And, voltage sensitivity for $M_1$ is given as:

$\text{V}_{s1}=\frac{\text{N}_{1}\text{B}_{1}\text{A}_{1}}{\text{K}_{1}\text{R}_{1}}$

$\therefore\text{Ratio}\frac{\text{V}_{\text{s}2}}{\text{V}_{\text{s}1}}=\frac{\text{N}_{2}\text{B}_{2}\text{A}_{2}\text{K}_{1}\text{R}_{1}}{\text{R}_{2}\text{K}_{2}\text{N}_{1}\text{B}_{1}\text{A}_{1}}$

$=\frac{42\times0.5\times1.8\times10^{-3}\times10\times\text{K}}{\text{K}\times14\times30\times0.25\times3.6\times10^{-3}}=1$
Hence, the ratio of voltage sensitivity of $M_2$ to $M_1$ is 1.

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Question 153 Marks

In a chamber, a uniform magnetic field of $6.5 \mathrm{G}\left(1 \mathrm{G}=10^{-4} \mathrm{~T}\right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
$\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{~m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$

Answer

Magnetic field strength, $B=6.5 \mathrm{G}=6.5 \times 10^{-4}$ TSpeed of the electron, $v=4.8 \times 10^6 \mathrm{~m} / \mathrm{s}$
Charge on the electron, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
Mass of the electron, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}$
Angle between the shot electron and magnetic field, $\theta=90^\circ$
Magnetic force exerted on the electron in the magnetic field is given as:
$\text{F}=\text{evB}\sin\theta$
This force provides centripetal force to the moving electron 1, Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
$\text{F}_{e}=\frac{\text{mv}^{2}}{\text{r}}$
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
$\text{F}_{e}=\text{F}$
$\frac{\text{mv}^{2}}{\text{r}}=\text{ev B}\sin\theta$
$=\frac{9.1\times10^{-31}\times4.8\times10^{6}}{6.5\times10^{-4}\times1.6\times10^{-19}\times\sin90^\circ}$
$=4.2\times10^{-2}\text{m}$
$=4.2\ \text{cm}$
Hence, the radius of the circular orbit of the electron is 4.2 cm.

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Question 163 Marks

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Answer

Resistance of the qalvanorneter coil, G = 15 Ω
Current for which there is full scale deflection,
$I_{g =}4 mA = 4 \times 10^{-3} A$
Range of the voltmeter is 0, which needs to be converted to 6 A.
$\therefore\text{Current, I}=6\text{A}$
a shunt resistor of resistance S is to be connected in parallel with the galvanorneter to convert it into a ammeter. This value of S is given as:
$\text{S}=\frac{\text{I}_{g}\text{G}}{\text{I}-\text{I}_{\text{g}}}$
$=\frac{4\times10^{-3}\times15}{6-4\times10^{-3}}$
$\text{S}=\frac{6\times10^{-2}}{6-0.004}=\frac{0.06}{5.996}$
≈ 0.01 Ω = 10 mΩ
Hence, a 10 mΩ shunt resistor is to be connected in parallel with the galvanometer.

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Question 173 Marks

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Answer

Length of the solenoid, I = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
$\therefore$ Total number of turns on the solenoid, N = 5 × 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
$\text{B}=\frac{\mu\text{NI}}{\text{l}}$
Where,
$\mu_{0}=$ Permeability of free space $=4\pi\times10^{-7}\ \text{Tm A}^{-1}$
$\text{B}=\frac{4\pi\times10^{-7}\times2000\times8}{0.8}$
$=8\pi\times10^{-3}=2.512\times10^{-2}\ \text{T}$
Hence, the magnitude of the magnetic field inside the solenoid near its centre is $2.512 \times 10^{-2}\ T.$

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Question 183 Marks

A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Answer

Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m
Magnitude of the magnetic field at this point is given as:
$\text{B}=\frac{\mu_{0}}{4\pi}\frac{2\text{l}}{\text{r}}$
Where,
$\mu_{0}$ = Permeability of free space
$=4\sqcap\times10^{-7}\text{T m A}^{-1}$
$\text{B}=\frac{4\pi\times10^{-7}\times2\times35}{4\pi\times0.2}$
$=3.5\times10^{-5}\text{T}$
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is $3.5 \times 10^{-5}\ T.$

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Question 193 Marks

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer

Length of the wire, I = 3 cm = 0.03 m
Current flowing in the wire, I = 10 A
Magnetic field, B = 0.27 T
Angle between the current and magnetic field, $\theta$ = 90°
Magnetic force exerted on the wire is given as:
$\text{f}=\text{BI}\ \sin\theta$
$=0.27\times10\times0.03\times\sin90^\circ$
$=8.1\times10^{-2}\ \text{N}$
Hence, the magnetic force per unit length on the wire is $8.1 \times 10^{-2}\ N$. The direction of the force can be obtained from Fleming's left hand rule.

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Question 203 Marks

Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Answer

Radius of coil $\mathrm{X}, \mathrm{r}_1=16 \mathrm{~cm}=0.16 \mathrm{~m}$
Radius of coil $\mathrm{Y}, \mathrm{r}_2=10 \mathrm{~cm}=0.1 \mathrm{~m}$
Number of turns of on coil $X, n_1=20$
Number of turns of on coil $\mathrm{Y}, \mathrm{n}_2=25$
Current in coil $X_1 I_1=16 \mathrm{~A}$
Current in coil $\mathrm{Y}, \mathrm{I}_2=18 \mathrm{~A}$
Magnetic field due to coil X at their centre is given by the relation,
$\text{B}=\frac{\mu\text{n}_{1}\text{I}_{1}}{2\text{r}_{1}}$
Where,
$\mu_{0}=\text{Permeability of free space}=4\pi\times10^{-7}\text{T m A}^{-1}$
$\therefore\text{B}_{1}=\frac{4\pi\times10^{-7}\times20\times16}{1\times0.16}$
$=4\pi\times10^{-4}\text{T}(\text{towards East})$
Magnetic field due to coil Y at their centre is given by the relation, $\text{B}=\frac{\mu\text{n}_{2}\text{I}_{2}}{2\text{r}_{2}}$
$\text{B}_{2}=\frac{4\pi\times10^{-7}\times25\times18}{2\times0.10}$
$=9\pi\times10^{-4}\text{T}(\text{towards East})$
Hence, net magnetic field can be obtained as:
$\text{B}=\text{B}_{2}-\text{B}_{1}$
$=9\pi\times10^{-4}-4\pi\times10^{-4}$
$=5\pi\times10^{-4}\text{T}$
$=1.57\times10^{-3}\text{T}(\text{towards west})$

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Question 213 Marks

A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is 60 cm, calculate the kinetic energy (in MeV) of the proton beam produced by the accelerator.

Answer

Magnetic field ? $ = 2\pi\text{mv}/\text{q}$
$ = \frac{2\times3.14 \times1.67\times 10^{−27}\times 10^{7}}{1.6\times 10^{−19}} = 0.66\text{T}$
Final velocity of proton $\text{v} =\text{R} \times2\pi\text{v} = 0.6\times 2 \times 3.14 \times 10^{7} = 3.77 \times 10^{7}\text{?/?}$
$\text{Energy} =\frac{1}{2}\text{mv}^{2}=\frac{1}{2}\times 1.67 \times 10^{−27}\times (3.77 \times 10^{7})^2\text{?} = 7.4\text{ ???}$.

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Question 223 Marks

Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R. Draw the magnetic field lines due to a circular wire carrying current I.

Answer

$\overrightarrow{\text{dB}} = \frac{\mu_{o}\overrightarrow{\text{dl}}\text{x}\overrightarrow{\text{r}}}{4\pi\text{r}^{3}}$

$\text{dB}_{x} = \frac{\mu_{o}\text{id}l\text{R}}{4\pi(\text{x}^{2} + \text{R}^{2})^{\frac{3}{2}}}$

$\overrightarrow{\text{B}} = \text{B}_{x}\hat{\text{i}} = \frac{\mu_{o}\text{IR}^{2}}{2(\text{x}^{2}\text{R}^{2})^{\frac{3}{2}}}\hat{\text{i}}$

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Question 233 Marks

Two identical loops P and Q each of radius 5 cm are lying in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils, if they carry currents equal to 3 A and 4 A respectively.

Answer

Field at the centre of a circular coil = $\frac{\mu _0{I}}{2R}$
Field due to coil $\text{P}=\frac{\mu_{0}\times3}{2\times5\times10^{-2}}\text{tesla}$
$=12\pi\times10^{-6}\text{tesla}$.
Field due to coil $\text{Q}=\frac{\mu_{0}\times4}{2\times5\times10^{-2}}\text{tesla}$

$=16\pi\times10^{-6}\text{tesla}$
$\therefore$ Resultant Field $=(\pi\sqrt{12^{2}+{16^{2}}})\mu\text{T}$

$=(20\pi)\mu{\text{T}}$
Let the field make an angle $\theta$ with the vertical
$\tan{\theta}=\frac{12\pi\times10^{-6}}{16\pi\times10{-6}}=\frac{3}{4}$
$\theta=\tan^{-1}\frac{3}{4}$

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Question 243 Marks

A cyclotron’s oscillator frequency is 10MHz. What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is 60 cm, calculate the kinetic energy (in MeV) of the proton beam produced by the accelerator.

Answer

Magnetic field $\text{B} = \frac{2\pi\text{mf}}{\text{q}}$
$ = \frac{2\times3.14 \times1.67\times 10^{−27}\times 10^{7}}{1.6\times 10^{−19}} = 0.66\text{T}$
Final velocity of proton $\text{v} =\text{R} \times2\pi\text{f} = 0.6\times 2 \times 3.14 \times 10^{7} = 3.77 \times 10^{7}\text{?/?}$
Energy $ =\frac{1}{2}\text{m}v^{2}=\frac{1}{2}\times 1.67 \times 10^{−27}\times (3.77 \times 10^{7})^2\text{?} = 7.4\text{ ???}.$

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Question 253 Marks

State Biot – Savart law and express this law in the vector form.
Two identical circular coils, P and Q each of radius R, carrying currents 1 A and $\sqrt{3}$ A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils.

Answer

It states that magnetic field strength, $\vec{dB}$ due to a current element, $I\vec{dl}$ at a point, having a position vector r relative to the current element is found to depend.

Directly on the current element.
Inversely on the square of the distance |r|.
Directly on the sine of angle between the current element and the position vector r.

In vector notation,

$\overrightarrow{dB}=\frac{\mu_0}{4\pi}\frac{I\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Alternate Answer

$\bigg(\overrightarrow{dB}=\frac{\mu_0}{4\pi}\frac{I\vec{dl}\times\hat{r}}{|\vec{r}|^2}\bigg)$

$B_p=\frac{\mu_0\times1}{2R}=\frac{\mu_0}{2R}$ (along z − direction)

$B_Q=\frac{\mu_0\times\sqrt{3}}{2R}=\frac{\mu_0\sqrt{3}}{2R}$ (along x − direction)

$\therefore\text{ }B=\sqrt{B_{p}\text{ }^2+B_Q\text{ }^2}=\frac{\mu_0}{R}$

This net magnetic field B, is inclined to the field Bp, at an angle $\theta$, where

$\tan\theta=\sqrt{3}$

$(/\theta=\tan^{-1}\sqrt{3}=60^\circ)$

(in XZ plane)

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Question 263 Marks

Define the current sensitivity of a galvanometer. Write its S.I. unit.
Figure shows two circuits each having a galvanometer and a battery of 3 V. When the galvanometers in each arrangement do not show any deflection, obtain the ratio $R_1/ R_2.$

Answer

Ratio of deflection produced in the galvanometer to the current flowing through it.
Current sensitivity $\text{S}_{i} = \frac{\theta}{\text{I}}$
S.I. unit of current sensitivity $S_i$ is division/ampere or radian/ampere.
For balanced Wheatone bridge, if no current flows through the galvanometer
$\frac{4}{\text{R}_{1}} = \frac{6}{9}\Rightarrow\text{R}_{1} = \frac{4\times9}{6} =6\Omega$
For another current
$\frac{6}{12} = \frac{\text{R}_{2}}{8}\Rightarrow\text{R}_{2} = \frac{6\times8}{12} = 4\Omega$
$\therefore\frac{\text{R}_{1}}{\text{R}_{2}} = \frac{6}{4} = \frac{3}{2}.$

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Question 273 Marks

A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the

Total torque on the coil.
Total force on the coil.
Average force on each electron in the coil, due to the magnetic field.

Assume the area of cross - section of the wire to be 10–5 m2 and the free electron density is $10^{29}/m^3.$

Answer

$\tau = \text{NBIA}\sin\theta$

$\text{ as }\theta = 0^{o}\text{ so }\tau = 0$

$\overrightarrow{\text{F}} = \text{I}(\overrightarrow{l}\times\overrightarrow{\text{B}})$

$\text{F}_{net} = 0$

$\text{F}_{e} = \text{evB}\text{ or } \text{F}_{e} = \frac{\text{BI}}{\text{NA}}$

$\text{F}_{e} = 1.5\times10^{-24}\text{N}.$

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Question 283 Marks

A proton and an $\alpha$ particle move perpendicular to a magnetic field. Find the ratio of radii of circular paths described by them when both have (i) equal velocities, and (ii) equal kinetic energy.

Answer

$r=\frac{mv}{qB}$

For proton $r_p=\frac{m_pv}{q_pB}$

For $\alpha$ particle $r_\alpha=\frac{m_{\alpha}v}{q_{\alpha}B}$

$\frac{r_p}{r_{\alpha}}=\frac{m_p}{q_p}\frac{q_{\alpha}}{m_{\alpha}}=\frac{1}{2}$

$r=\frac{\sqrt{2mk}}{q_B}$

$r_{p}=\frac{\sqrt{2m_{p}K}}{q_{p}B}$

$r_\alpha=\sqrt{2m_\alpha{k}} \over{q_\alpha}B$

$\frac{r_p}{r{\alpha}}=\frac{q_{\alpha}}{q_p}\sqrt{\frac{m_p}{m_{\alpha}}}=\frac{1}{1}$

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Question 293 Marks

A point charge q moving with speed v enters a uniform magnetic field B that is acting into the plane of the paper as shown. What is the path followed by the charge q and in which plane does it move?
How does the path followed by the charge get affected if its velocity has a component parallel to $\overrightarrow{\text{B}}$?
If an electric field $\overrightarrow{\text{E}}$ is also applied such that the particle continues moving along the original straight line path, what should be the magnitude and direction of the electric field $\overrightarrow{\text{E}}$?

Answer

The charge q describes a circular path; anti-clockwise in XY plane.
The path will become helical.
Direction of Lorentz magnetic force is –Y

$\therefore$ Applied electric field should be in +Y direction.

$F_E=F_m$

$\Rightarrow qE=qvB$

$\Rightarrow E=vB$

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Question 303 Marks

The figure shows three infinitely long straight parallel current carrying conductors. Find the

magnitude and direction of the net magnetic field at point A lying on conductor 1,
magnetic force on conductor 2.

Answer

$B_2=\frac{\mu_0}{4\pi}\frac{2(3I)}{r}=\frac{\mu_0}{4\pi}\big(\frac{6I}{r}\big)$into the plane of the paper/$(\otimes)$.

$B_3=\frac{\mu_0}{4\pi}\frac{2(4I)}{3r}=\frac{\mu_0}{4\pi}\big(\frac{8I}{3r}\big)$out of the plane of the paper/$(\bigodot)$.

$B_A=B_2-B_3$ into the paper.

$=\frac{\mu_0}{4\pi}\Big(\frac{10I}{3r}\Big) into\text{ }the\text{ }paper.\text{} (\otimes)$

$F_{21}=\frac{\mu_0}{4\pi}\frac{2I(3I)}{r}$ away from wire1 (/towards 3)

$F_{23}=\frac{\mu_0}{4\pi}\frac{2(3I)(4I)}{2r}$ away from wire 3 (towards 1)

$F_{\text{net}}=F_{23}-F_{21}$ towards wire1

$=\frac{\mu_0}{4\pi}\frac{6(I)^2}{r}$ towards wire1

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Question 313 Marks

Using Biot-Savart law, deduce the expression for the magnetic field at a point (x) on the axis of a circular current carrying loop of radius R. How is the direction of the magnetic field determined at this point?

Answer

$\overrightarrow{dB}=\frac{\mu_0}{4\pi}\text{ }\text{I}\text{ }\frac{\vec{dl}\times\overrightarrow{r}}{r^3}$
$[\text{OR}\text{ }dB=\frac{\mu_0}{4\pi}\frac{Idl}{r^2}]$
$\text{Here}\text{ }r^2=x^2+R^2$
$\text{dB}=\frac{\mu_0}{4\pi}\frac{I dl}{x^2+R^2}$
$\sum\text{dB}_\perp=0$
$\text{d}B_x=\text{dB}\cos\theta\text{ }\text{ }\text{ }\text{ }\text{w}here\text{ }\text{ }\text{ }\cos\theta=\frac{R}{(x^2+R^2)^{\text{ }^1/_2}}$
$\text{d}B_x=\frac{\mu_0\text{ }Idl}{4\pi}\frac{R}{(x^2+R^2)^{\text{ }^3/_2}}$
$\overrightarrow{B}=\int\text{d}b_x\hat{i}=\frac{\mu_0IR^2}{2(x^2+R^2)^{\text{ }^3/_2}}\hat{i}$

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Question 323 Marks

State the condition under which a charged particle moving with velocity v goes undeflected in a magnetic field B.
An electron, after being accelerated through a potential difference of 104 V, enters a uniform magnetic field of 0.04 T, perpendicular to its direction of motion. Calculate the radius of curvature of its trajectory.

Answer

The force experienced $\vec{F}=q(\vec{v}\times\vec{B})$

The charge will go undeflected when $\vec{v}$ is parallel or

antiparallel to$\vec{B}\because\vec{F}=0$

The radius of electron

$eV=\frac{1}{2}mv^2$

$\frac{mv^2}{r}=qvB$

$\therefore\text{ }r=\frac{1}{B}\sqrt{\frac{2mV}{e}}$

$=\Bigg[\sqrt{\frac{2\times9.1\times10^{-31}\times10^4}{1.6\times10^{-19}}}\times\frac{1}{0.04}\Bigg]m$

$=8.4\times10^{-3}m$

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Question 333 Marks

The figure shows three infinitely long straight parallel current carrying conductors. Find the:

Magnitude and direction of the net magnetic field at point A lying on conductor 1.
Magnetic force on conductor 2.

Answer

$B_2=\frac{\mu_0}{4\pi}\frac{2(3I)}{r}=\frac{\mu_0}{4\pi}\big(\frac{6I}{r}\big)$into the plane of the paper/$(\otimes)$.

$B_3=\frac{\mu_0}{4\pi}\frac{2(4I)}{3r}=\frac{\mu_0}{4\pi}\big(\frac{8I}{3r}\big)$out of the plane of the paper/$(\bigodot)$.

$B_A=B_2-B_3$ into the paper.

$=\frac{\mu_0}{4\pi}\Big(\frac{10I}{3r}\Big) into\text{ }the\text{ }paper.\text{} (\otimes)$

$F_{21}=\frac{\mu_0}{4\pi}\frac{2I(3I)}{r}$ away from wire1 (/towards 3)

$F_{23}=\frac{\mu_0}{4\pi}\frac{2(3I)(4I)}{2r}$ away from wire 3 (towards 1)

$F_{\text{net}}=F_{23}-F_{21}$ towards wire1

$=\frac{\mu_0}{4\pi}\frac{6(I)^2}{r}$ towards wire1

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Question 343 Marks

Two identical coils P and Q each of radius R are lying in perpendicular planes such that they have a common centre. Find the magnitude and direction of the magnetic field at the common centre of the two coils, if they carry currents equal to $\text{I}$ and $\sqrt{3\text{I}}$ respectively.

Answer

$B_p=\frac{\mu_0}{4\pi}\frac{2\pi I}{R}$
$B_Q=\frac{\mu_0}{4\pi}\frac{2\pi(\sqrt{3}I)}{R}$
$B=\sqrt{B_P\text{ }^2+B_Q\text{ }^2}$
$=\frac{\mu_0}{4\pi}\frac{2\pi I}{R}\sqrt{1+3}$
$=\frac{\mu_0I}{R}$
$\tan\theta=\frac{B_p}{B_Q}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=30^\circ$

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Question 353 Marks

Write the expression for the force $ \overrightarrow{\text{F}}$ acting on a particle of mass m and charge q moving with velocity $ \overrightarrow{\text{V}} $ in a magnetic field $ \overrightarrow{\text{B}}$ Under what conditions will it move in (i) a circular path and (ii) a helical path?
Show that the kinetic energy of the particle moving in a magnetic field remains constant.

Answer

$ \overrightarrow{\text{F}}=Q(\overrightarrow{\text{V}}\times\overrightarrow{\text{B}})$

When velocity of charged particle and magnetic field are perpendicular to each other.
When velocity is neither parallel nor perpendicular to the magnetic field.

The force, experienced by the charged particle, is perpendicular to the instantaneous velocity $\overrightarrow{V}$ at all instants. Hence the magnetic force cannot bring any change in the speed of the charged particle. Since speed remains constant, the kinetic energy also stays constant.

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Question 363 Marks

Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer.
Can a galvanometer as such be used for measuring the current? Explain.

Answer

When a coil, carrying current, and free to rotate about a fixed axis, is placed in a uniform magnetic field, it experiences a torque (which is balanced by a restoring torque of suspension).
To have deflection proportional to current/to maximize the deflecting torque acting on the current carrying coil.
To make magnetic field radial/to increase the strength of magnetic field.
Expression for current sensitivity
$I_S=\frac{\theta}{I}\text{ }or\text{ }\frac{NAB}{K}$
where $\theta$ is the deflection of the coil
No
The galvanometer, can only detect currents but cannot measure them as it is not calibrated. The galvanometer coil is likely to be damaged by currents in the (mA/A) range.

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Question 373 Marks

Two long straight parallel conductors carry steady current $I_1$ and $I_2$ separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.

Answer

As shown in Figure, the direction of force on conductor b is attractive.
Alternate Answer
$\overrightarrow{\text{B}}$at a point on wire 2, is along - $\hat{\text{k}}$
$\therefore\overrightarrow{\text{F}} , $ on wire 2, due to the $\overrightarrow{\text{B}}$ is along - $\hat{\text{i}}$ i.e.towards wire1. Hence the force is attractive.

Magnetic field, due to current in conductor a,
$\text{B}_{1} = \frac{\mu_\circ\text{I}_{1}}{2\pi\text{d}}$
The magnitude of force on a length L of conductor b,
$\text{F}_{2} = \text{I}_{2}\text{LB}_{1}$
$\text{F}_{2} =\frac{\mu_\circ\text{I}_{1}\text{I}_{2}\text{L}}{2\pi\text{d}}$
One ampere is that steady current which, when maintained in each of the two very long, straight, parallel conductors, placed one meter apart in vacuum, would produce on each of these conductors a force equal to $2 \times10^{-7}$ newton per meter of their length.

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Question 383 Marks

Write the expression for the magnetic force acting on a charged particle moving with velocity n in the presence of magnetic field B. A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer.

Answer

$\overrightarrow{\text{F}} = \text{q}(\overrightarrow{\text{v}}\times\overrightarrow{\text{B}})$
$\text{F}= \text{qvB}\sin\theta$ and Force (F) acts perpendicular to the plane containing $\overrightarrow{\text{v}} \text{and} \overrightarrow{\text{B}}$

Justification: Direction of force experienced by the particle will be according to the Fleming’s Left-hand rule.

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Question 393 Marks

A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor.

Answer

Energy stored, in the capacitor of capacitance 12 pF,
$\text{U}=\frac{1}{2}\text{CV}^2$
$=\frac{1}{2}\times12\times10^{-12}\times50\times50\text{J}$
$=1.5\times10^{-8}\text{ }\text{J}$
C= Equivalent capacitance of 12 pF and 6 pF, in series, is given by
$\frac{1}{\text{C}}=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}$
$\therefore\text{C}=4\text{ }\text{pF}$
$\therefore$ Charge stored across each capacitor
$\text{q}=\text{CV}$
$=4\times10^{-12}\times50\text{ }\text{C}$
$=2\times10^{-10}\text{ }\text{C}$
Charge on each capacitor 12 pF as well as 6 pF
$\therefore$ Potential difference across capacitor $C_1$
$\therefore\text{V}_{1}=\frac {2\times10^{-10}}{12\times10^{-12}}\text{Volt}=\frac{50}{3}\text{ }\text{V}$
Potential difference across capacitor $C_2$
$\text{V}_{2}=\frac {2\times10^{-10}}{6\times10^{-12}}\text{Volt}=\frac{100}{3}\text{ }\text{V}$

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Question 403 Marks

A electron of mass $m_e$ revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated wit it is expressed as $\vec\mu=-_{e2{\text{ me}}}\overrightarrow{\text{L}}$, where $\overrightarrow{\text{L}}$ is the orbital angular momentum of the electron. Give the significance of negative sign.

Answer

Electron, in circular motion around the nucleus, constitutes a current loop which behaves like a magnetic dipole. Current associated with the revolving electron: $I=\frac{e}{T}$ $\text{and}\text{ }\text{T}=\frac{2{\pi{r}}}{\vartheta}$
$\therefore\text{I}=\frac{e}{2{\pi}r}\vartheta$

Magnetic moment of the loop, $\mu=\text{IA}$
$\mu=\text{IA}=\frac{e\vartheta}{2{\pi}r}\pi{r}^2=\frac{{e\vartheta}r}{2}=\frac{e.m_{e}\vartheta{r}}{2m_{e}}$ Orbital angular momentum of the electron, $ \text{L}= m_e\vartheta{r}$ $\vec\mu=\frac{-e}{2{m}_e}\vec{L}$
-ve sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.

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Question 413 Marks

A rectangular loop of wire of size 4cm × 10cm carries a steady current of 2A. A straight long wire carrying 5A current is kept near the loop as shown. If the loop and the wire are coplanar, find

The torque acting on the loop and.
The magnitude and direction of the force on the loop due to the current carrying wire.

Answer

Torque on the loop

$\tau =\text{MB}\sin\theta$
As Mand B are parallel, $\theta =0 $
therefore, $\tau = 0 $

Force acting on the loop

$|\text{F}| =\frac{\mu_{o}\text{I}_{1}\text{I}_{2}}{2\pi}\ell\bigg(\frac{1}{\text{r}_{1}} -\frac{1}{\text{r}_{2}}\bigg)$
$ = 2\times10^{-7}\times2\times2\times10^{-1}\bigg(\frac{1}{10^{-2}}-\frac{1}{5\times10^{-2}}\bigg)\text{N}$
$ =\frac{8\times10^{8}}{10^{-2}}\bigg(1 - \frac{1}{5}\bigg)\text{N}$
$ = 8\times10^{-6}\bigg(\frac{4}{5}\bigg)\text{N}$
$ = 6.4\times10^{-6}\text{N}$
Direction: Towards conductor /Attractive
Magnetic field $\text{B} =\frac{\mu_{0}\text{I}}{2\pi\text{r}}$
$|\text{F}_{1}| =\text{I}\ell\text{B}_{1}\text{ (towards the conductor) }$
$|\text{F}_{2}| =\text{I}\ell\text{B}_{2}\text{ (away from the conductor}$
$|\text{F}| = \text{I}\ell(\text{B}_{1} - \text{B}_{2} )$
$ = \text{I}\ell\bigg(\frac{\mu_{0}\text{i}}{2\pi\text{r}_{1}} -\frac{\mu_{0}\text{i}}{2\pi\text{r}_{2}}\bigg) = \frac{\mu_{0}\text{i }l\text{I}}{2\pi}\bigg(\frac{1}{\text{r}_{1}} - \frac{1}{\text{r}_{2}}\bigg)$
$|\text{F}| = \frac{2\times10^{-7}\times(2)^{2}\times10^{-1}}{10^{-2}}\bigg(1 - \frac{1}{5}\bigg)\text{N}$
$ = 6.4\times10^{-6}\text{N}$
Direction: Towards the conductor/attractive.

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Question 423 Marks

A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across the cross-section. Apply Ampere’s circuital law to calculate the magnetic field at a point ‘r’ in the region for (i) r < a and (ii) r > a.

Answer

Consider the case r < a. The Amperian loop is a circle labelled 1. For this loop, taking the radius of the circle to be r, L = 2 $\pi\text{r}.$

Now the current enclosed $\text{I}_{e} = \text{I}\bigg(\frac{\pi\text{r}^{2}}{\pi\text{a}^{2}}\bigg) = \frac{\text{Ir}^{2}}{\text{a}^{2}}$

Using Ampere’s law,

$\text{B}(2\pi\text{r}) = \frac{\mu_{o}\text{Ir}^{2}}{\text{a}^{2}}\Rightarrow\text{B} = \frac{\mu_{o}\text{Ir}^{2}}{2\pi\text{a}^{2}}$

Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop, L = 2 $\pi\text{r}.$

Ie = Current enclosed by the loop = I

$\text{B}(2\pi\text{r}) = \mu_{o}\text{I}$

$\Rightarrow\text{B} =\frac{\mu_{o}\text{I}}{2\pi\text{r}}.$

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Question 433 Marks

Write the expression for the magnetic moment ($\overrightarrow{\text{m}}$) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form. In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance l as shown. Give reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.

Answer

$\overrightarrow{\text{m}} = \text{I}\overrightarrow{\text{A }} ( \overrightarrow{\text{A}} = \text{ area vector) }$
Torque: The magnetic field due to the long current carrying wire is perpendicular to the plane of paper. Hence the force acting on each of the four sides is in the plane of the paper and the net torque is zero.
Alternate Answer
$\overrightarrow{\text{m}}$is perpendicular to the plane of paper and$\overrightarrow{\text{B}}$ is perpendicular to the plane of paper. Hence $\overrightarrow{\tau} =\overrightarrow{\text{m}}\times\overrightarrow{\text{B}} = 0 $
Force:
Force on upper horizontal side $ = \frac{\text{I}l\mu_{o}\text{I}_{1}}{2\pi\text{l}} =\frac{\text{I}\mu_{o}\text{l}_{1}}{2\pi}\text{ (attractive) }$
Force on lower horizontal side $ = \frac{\text{I}l\mu_{o}\text{I}_{1}}{2\pi(2l)} = \frac{\text{I}\mu_{o}\text{I}_{1}}{4\pi}\text{ (repulsive)}$
The direction of these forces being opposite to each other therefore net force $ = \frac{\mu_{o}\text{I}_{1}\text{I}}{4\pi}\text{ (attractive) }$
(The net force on the two vertical sides is zero)
Torque: The magnetic field due to the long current carrying wire is perpendicular to the plane of paper. Hence the force acting on each of the four sides is in the plane of the paper and the net torque is zero.

Alternate Answer
$\overrightarrow{\text{m}}$is perpendicular to the plane of paper and$\overrightarrow{\text{B}}$ is perpendicular to the plane of paper. Hence $\overrightarrow{\tau} =\overrightarrow{\text{m}}\times\overrightarrow{\text{B}} = 0 $
Force: Award this mark irrespective of result obtained or calculation done by the students.

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Question 443 Marks

State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer can not be used as such to measure current in a given circuit. Name any two factors on which the current sensitivity of a galvanometer depends.

Answer

Principle: Torque acts on a current carrying coil suspended in magnetic field.
$(\tau =\text{NIAB}\sin\theta)$
Two reasons:

Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of a few μA.
For measuring currents, the galvanometer has to be connected in series, and as it has a finite resistance, this will change the value of the current in the circuit.

Two factors: The current sensitivity of a moving coil galvanometer can be increased by (i) increasing the number of turns (ii) increasing area of the loop. (iii) increasing magnetic field (iv) decreasing the torsional constant of the suspension wire.

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Question 453 Marks

Derive the expression for force per unit length between two long straight parallel current carrying conductors. Hence define one ampere.

Answer

Figure shows two long parallel conductors a and b separated by a distance d and carrying (parallel) currents $I_a$ and $I _b,$ respectively. The conductor ‘a’ produces, the same magnetic field $B_a$ at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). Its magnitude is given by
$\text{B}_{\alpha} = \frac{\mu_\circ\text{I}_{\alpha}}{2\pi\text{d}}$
The conductor ‘b’ carrying a current $I _b$ will experience a sideways force due to the field $B_a$. The direction of this force is towards the conductor ‘a’. We label this force as $F_{ba}$, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by
$\int_{ba} = \text{I}_{b}\text{LB}_{a}$ $ = \frac{\mu_\circ\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}\text{L}$
Let $\int_{ba}$ represent the magnitude of the force $\text{F}_{ba}$ per unit length.
$\int_{ba} = \frac{\mu_\circ\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}$
One ampere: The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to $2 × 10^{-7}$ newton per metre of their length.

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Question 463 Marks

Explain the principle and working of a cyclotron with the help of a schematic diagram. Write the expression for cyclotron frequency.

Answer

Principle: When charged particles are passed through a suitable combination of crossed electric and magnetic fields there energy can increase.
Labelled diagram:

Working: Inside the metal boxes the particle is shielded and is not acted on by the electric field. The magnetic field, however, acts on the particle and makes it go round in a circular path inside a dee. Every time the particle moves from one dee to another it is acted upon by the electric field. The sign of the electric field is changed alternately in tune with the circular motion of the particle. This ensures that the particle is always accelerated by the electric field. Each time the acceleration increases the energy of the particle.
$\tau = \frac{1}{\text{v}_{c}} = \frac{2\pi\text{m}}{\text{qB}}$
or $\text{v}_{c} = \frac{\text{qB}}{2\pi\text{m}}$.

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Question 473 Marks

A bar magnet of magnetic moment 6J/T is aligned at 60° with a uniform external magnetic field of 0·44T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Answer

$\text{m}=6\frac{\text{J}}{\text{T}}$

$\theta_1=60^\circ$

$\text{B}=0.44\text{T}$

$\text{W}=-\text{mB}[\cos\theta_2-\cos\theta_1]$

$\theta_2=90^\circ$

$\text{W}=-6\times0.44[\cos90^\circ-\cos60^\circ]$

$=-2.64\Big[0-\frac{1}{2}\Big]$

$\text{W}=1.32\text{J}$

$\theta_2=180^\circ$

$\text{W}=-6\times0.44[\cos180^\circ-\cos60^\circ]$

$=-2.64\Big[-1-\frac{1}{2}\Big]$

$\text{W}=2.64\times\frac{3}{2}$

$\text{W}=3.965\text{J}$

When $\theta=180^\circ$

Torque will be

$\tau=\text{mB}\sin\theta$

$=6\times0.44\times\sin180^\circ$

$\tau=0\text{Nm}$

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Question 483 Marks

Write an expression of magnetic moment associated with a current (I) carrying circular coil of radius r having N turns.
Consider the above mentioned coil placed in YZ plane with its centre at the origin. Derive expression for the value of magnetic field due to it at point (x, 0, 0).

Answer

Magnetic moment $\vec{\text{M}}=\text{Ni}(\pi\text{r}^2)\hat{\text{n}}.$

Magnetic field at point P(x, 0, 0) due to $\vec{\text{Idl}}$
$\vec{\text{dB}}=\frac{\mu_0}{4\pi}\frac{\text{Idl}\sin90^\circ}{\text{r}^2_1}\text{along PQ}$
For entire coil $\int\vec{\text{dB}}\cos\theta=0$
$\therefore\vec{\text{B}}\text{ at }\text{P}$
$\Rightarrow\text{B}=\int\text{dB}\sin\theta=\frac{\mu_0\text{I}\sin\text{R}}{4\pi\text{R}^2}\int\limits^{2\pi\text{r}}_{0}\text{dl}$
$=\frac{\mu_0\text{I}}{4\pi\text{r}^2_1}\times\frac{\text{r}}{\text{r}_1}\times(2\pi\text{r})$
$\Rightarrow\vec{\text{B}}=\frac{\mu_0\text{Ir}^2}{2\big(\text{r}^2+\text{x}^2\big)^\frac{3}{2}}\hat{\text{i}}$
Coil has N turns then,
$\vec{\text{B}}=\frac{\mu_0\text{INr}^2}{2\big(\text{r}^2+\text{x}^2\big)^\frac{3}{2}}\hat{\text{i}}$

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Question 493 Marks

Define current sensitivity of a galvanometer. Write its expression.
A galvanometer has resistance G and shows full scale deflection for current Ig.

How can it be converted into an ammeter to measure current up to $I_0(I_0 > Ig)?$
What is the effective resistance of this ammeter?

Answer

Current sensitivity: It is defined as the amount of deflection produced per unit magnitude of current passes.

$\text{C}_\text{s}=\frac{\phi}{\text{I}}\text{ or }\text{C}_\text{s}=\frac{\text{NAB}}{\mu,}$

(G) can be converted into an ammeter by connected a small stunt resistance parallel to (G) coil so that,
$\text{IgG}=\big(\text{I}_0-\text{I}_\text{g}\big)\text{S}$
$\therefore\text{S}=\frac{\text{IgG}}{\text{I}_0-\text{I}_\text{g}}$

Effective resistance of $(\text{A})\Rightarrow\frac{\text{GS}}{\text{G}+\text{S}}$

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Question 503 Marks

Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.
Explain the significance of a radial magnetic field when a current carrying coil is kept in it.

Answer

The plane of the loop is not along the magnetic field but makes an angle with it.

Let the dimension of the rectangular coil ABCD, be AB × BC = a × b
The angle between the field and the normal is θ.
Forces on BC and DA are equal and opposite and they cancel each other as they are collinear.
Force on AB is $F_1$ and force on CD is $F_2.$
$F_1 = F_2 = IbB$
The magnitude of the torque on the loop as in the figure:

$\therefore\tau=\text{F}_1\frac{\text{a}}{2}\sin\theta+\text{F}_2\frac{\text{a}}{2}\sin\theta$
$=\text{lab}\text{B}\sin\theta$
$\tau=\text{lab}\sin\theta$
If there are 'n' such turns the torque will be $\text{n}\text{lab}\sin\theta$
The magnetic moment of the current, m = lA
$\therefore\overrightarrow{\tau}=\overrightarrow{\text{m}}\times\overrightarrow{\text{B}}$

The uniform radial magnetic field keeps the plane of the coil always parallel to the direction of the magnetic field. that is, the angle between the plane of the coil and the magnetic field is zero in all the orientation of the coil.

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