Question 12 Marks
Draw a diagram of magnetic field lines due to any current carrying circular loop.
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Question 22 Marks
Radii of two circular current carrying coils are $r_1$ and $r_2$ respectively and number of turns in them are $n_1$ and $n_2$ respectively. They are joined in series. Prove that the ratio of produced magnetic fields at their centers is $n_1 r_2: n_2 r_1$. If coils are joined in parallel, then the ratio of magnetic fields produced at the centers is equivalent to $\left(r_2 / r_1\right)^2$.
Answer
View full question & answer→Since the coils are joined in series, same current I will flow through them. Therefore, magnetic fields at the centers are
$B _1=\frac{\mu_0 n_1 I }{2 r_1}$ and $B _2=\frac{\mu_0 n_2 I }{2 r_2}$
$\therefore \frac{ B _1}{B_2}=\frac{\frac{\mu_0 n_1 I }{2 r_1}}{\frac{\mu_0 n_2 I }{2 r_2}}$
$\Rightarrow \quad \frac{ B _1}{B_2}=\frac{n_1}{r_1} \times \frac{r_2}{n_2}=\frac{n_1 r_2}{r_1 n_2}$
$\Rightarrow \quad B _1: B _2=n_1 r_2: n_2 r_1$ Ans.
When the coils are joined in parallel, then the ratio of currents in them is equal to reciprocal of ratio of their resistances $R _1$ and $R _2$, i.e., $\frac{ I _1}{ I _2}=\frac{ R _2}{ R _1}=\frac{n_2\left(2 \pi r_2\right) \rho}{n_1\left(2 \pi r_1\right) \rho}=\frac{n_2 r_2}{n_1 r_1}$
$\therefore \quad \frac{ B _1}{B_2}=\frac{\mu_0 n_1 I _1}{2 r_1} \times \frac{2 r_2}{\mu_0 n_2 I _2}$
$=\frac{n_1 r_2}{n_2 r_1} \times \frac{n_2 r_2}{n_1 r_1}=\frac{r_2^2}{r_1^2}$
$B _1=\frac{\mu_0 n_1 I }{2 r_1}$ and $B _2=\frac{\mu_0 n_2 I }{2 r_2}$
$\therefore \frac{ B _1}{B_2}=\frac{\frac{\mu_0 n_1 I }{2 r_1}}{\frac{\mu_0 n_2 I }{2 r_2}}$
$\Rightarrow \quad \frac{ B _1}{B_2}=\frac{n_1}{r_1} \times \frac{r_2}{n_2}=\frac{n_1 r_2}{r_1 n_2}$
$\Rightarrow \quad B _1: B _2=n_1 r_2: n_2 r_1$ Ans.
When the coils are joined in parallel, then the ratio of currents in them is equal to reciprocal of ratio of their resistances $R _1$ and $R _2$, i.e., $\frac{ I _1}{ I _2}=\frac{ R _2}{ R _1}=\frac{n_2\left(2 \pi r_2\right) \rho}{n_1\left(2 \pi r_1\right) \rho}=\frac{n_2 r_2}{n_1 r_1}$
$\therefore \quad \frac{ B _1}{B_2}=\frac{\mu_0 n_1 I _1}{2 r_1} \times \frac{2 r_2}{\mu_0 n_2 I _2}$
$=\frac{n_1 r_2}{n_2 r_1} \times \frac{n_2 r_2}{n_1 r_1}=\frac{r_2^2}{r_1^2}$
Question 32 Marks
Resistance of a galvanometer is $30 \Omega$. A current of 2 mA gives full scale deflection in it. Calculate the necessary resistance required to make it an ammeter of range $(0-0.3 A)$.
Answer
View full question & answer→Sol. Given: $\quad G =30 \Omega$
$I _g=2 mA=2 \times 10^{-3} A$
and $I=0.3 A $
Shunt resistance to be joined in parallel with it to make it an ammeter in the range $(0-0.3 A)$
$S =\frac{ I _g}{\left( I - I _g\right)} \cdot G$
$=\frac{\left(2 \times 10^{-3}\right) \times 30}{\left(0.3-2 \times 10^{-3}\right)}$
$=\frac{60 \times 10^{-3} \times 10^3}{(300-2)}=\frac{60}{298}$
$=0.20 \Omega$ Ans.
$I _g=2 mA=2 \times 10^{-3} A$
and $I=0.3 A $
Shunt resistance to be joined in parallel with it to make it an ammeter in the range $(0-0.3 A)$
$S =\frac{ I _g}{\left( I - I _g\right)} \cdot G$
$=\frac{\left(2 \times 10^{-3}\right) \times 30}{\left(0.3-2 \times 10^{-3}\right)}$
$=\frac{60 \times 10^{-3} \times 10^3}{(300-2)}=\frac{60}{298}$
$=0.20 \Omega$ Ans.
Question 42 Marks
In a solenoid, there are 500 turns/meter and current flowing in it is 5 A . Length of the solenoid is 0.5 m and its radius is 1 cm . Calculate the intensity of magnetic field inside the solenoid.
Answer
View full question & answer→Here the number of turns of solenoid per unit length, $n=500$ per meter
Current in the solenoid, $I =5 A$
$\therefore$ Intensity of magnetic field inside the solenoid $ B =\mu_0 n I $
or $B =\left(\frac{\mu_0}{4 \pi}\right), 4 \pi n I$
$=\frac{4 \pi \times 10^{-7} \times 4 \times 3.14 \times 500 \times 5}{4 \pi}$
$=3.14 \times 10^{-3}$ Weber $/$ meter $^2$ Ans.
Note: In the above formula for B, there is no need of length and radius of the solenoid.
Current in the solenoid, $I =5 A$
$\therefore$ Intensity of magnetic field inside the solenoid $ B =\mu_0 n I $
or $B =\left(\frac{\mu_0}{4 \pi}\right), 4 \pi n I$
$=\frac{4 \pi \times 10^{-7} \times 4 \times 3.14 \times 500 \times 5}{4 \pi}$
$=3.14 \times 10^{-3}$ Weber $/$ meter $^2$ Ans.
Note: In the above formula for B, there is no need of length and radius of the solenoid.
Question 52 Marks
A coil of N turns is wound tightly in the form of a spiral whose internal and external radii $r_1$ are $r_2$ and respectively. Current $I$ is flowing through the coil, then calculate the magnetic field at the center.
Answer
View full question & answer→Suppose there is a loop of very small thickness $\delta r$ of the arbitrary radius $r$ of the spiral coil, then due to this loop the magnetic field at the center is
$\delta B =\frac{\mu_0(\delta N) I }{2 r}$
$\delta N =\frac{ N }{\left(r_2-r_1\right)} \times \delta r$
$\therefore \quad \delta B =\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \times \frac{\delta r }{r}$
On integration,
$B =\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \int_{r_1}^{r_2} \frac{d r}{r}$
$=\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \times[\ln r]_{r_1}^{r_2}$
$=\frac{\mu_0 NI }{2\left(r_2-r_1\right)}\left[\ln r_2-\ln r_1\right]$
or $B =\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \ln \frac{r_2}{r_1}$ Ans.
$\delta B =\frac{\mu_0(\delta N) I }{2 r}$
$\delta N =\frac{ N }{\left(r_2-r_1\right)} \times \delta r$
$\therefore \quad \delta B =\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \times \frac{\delta r }{r}$
On integration,
$B =\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \int_{r_1}^{r_2} \frac{d r}{r}$
$=\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \times[\ln r]_{r_1}^{r_2}$
$=\frac{\mu_0 NI }{2\left(r_2-r_1\right)}\left[\ln r_2-\ln r_1\right]$
or $B =\frac{\mu_0 NI }{2\left(r_2-r_1\right)} \ln \frac{r_2}{r_1}$ Ans.
Question 62 Marks
Two identical coils, radius of each is 8 cm and number of turns is 100 are coaxial and their centers are 12 cm apart. If current of 1 A flows in each coil in the same direction, then calculate the magnetic field at the midpoint on the axial line.
Answer
View full question & answer→Given, $a=8 cm=0.08 m, n=100, x=6 cm=0.06 m$
Produced magnetic field at a distance of 0.06 m from center of one coil
$ B_1=\frac{\mu_0 n l a^2}{2\left(a^2+x^2\right)^{3 / 2}} $
On putting the values,
$B _1=\frac{4 \pi \times 10^{-7} \times 100 \times 1 \times(0.08)^2}{2\left[(0.08)^2+(0.06)^2\right]^{3 / 2}}$
$=\frac{12.56 \times 10^{-7} \times 10^2 \times 64 \times 10^{-4}}{2\left[64 \times 10^{-4}+36 \times 10^{-4}\right]^{3 / 2}}$
$=\frac{12.56 \times 64 \times 10^{-9}}{2\left[100 \times 10^{-4}\right]^{3 / 2}}=\frac{12.56 \times 32 \times 10^{-9}}{1000 \times 10^{-6}}$
$=\frac{401.92 \times 10^{-9}}{10^{-3}}=4.02 \times 10^{-4} Tesla$
Since current is flowing in the same direction in both the coils, so magnitude of magnetic field on the axial line at the midpoint is
$ \begin{aligned} B & =2 B_1=2 \times 4.02 \times 10^{-4} \\ & =8.04 \times 10^{-4} Tesla \end{aligned} $
Produced magnetic field at a distance of 0.06 m from center of one coil
$ B_1=\frac{\mu_0 n l a^2}{2\left(a^2+x^2\right)^{3 / 2}} $
On putting the values,
$B _1=\frac{4 \pi \times 10^{-7} \times 100 \times 1 \times(0.08)^2}{2\left[(0.08)^2+(0.06)^2\right]^{3 / 2}}$
$=\frac{12.56 \times 10^{-7} \times 10^2 \times 64 \times 10^{-4}}{2\left[64 \times 10^{-4}+36 \times 10^{-4}\right]^{3 / 2}}$
$=\frac{12.56 \times 64 \times 10^{-9}}{2\left[100 \times 10^{-4}\right]^{3 / 2}}=\frac{12.56 \times 32 \times 10^{-9}}{1000 \times 10^{-6}}$
$=\frac{401.92 \times 10^{-9}}{10^{-3}}=4.02 \times 10^{-4} Tesla$
Since current is flowing in the same direction in both the coils, so magnitude of magnetic field on the axial line at the midpoint is
$ \begin{aligned} B & =2 B_1=2 \times 4.02 \times 10^{-4} \\ & =8.04 \times 10^{-4} Tesla \end{aligned} $
Question 72 Marks
In a long straight wire, a current of 5 A is flowing. How much magnetic field will be produced at a distance of 10 cm from it? Which rule will give the direction of the field?
Answer
View full question & answer→Given : $I =5 A, r=10 cm=0.10 m$
Produced magnetic field $\vec{B}$ on the observation point due to this
$ B=\frac{\mu_0}{4 \pi}\left(\frac{2 I}{r}\right) $
On putting the values, $=10^{-7}\left(\frac{2 I }{r}\right)$
$B =10^{-7}\left(\frac{2 \times 5}{0.10}\right)$
$=10^{-5} N / A - m$ Ans
Direction of the field $\vec{B}$ can be determined by rule number 1 of right-hand palm rule, or Maxwell cork screw rule or right-hand thumb rule.
Produced magnetic field $\vec{B}$ on the observation point due to this
$ B=\frac{\mu_0}{4 \pi}\left(\frac{2 I}{r}\right) $
On putting the values, $=10^{-7}\left(\frac{2 I }{r}\right)$
$B =10^{-7}\left(\frac{2 \times 5}{0.10}\right)$
$=10^{-5} N / A - m$ Ans
Direction of the field $\vec{B}$ can be determined by rule number 1 of right-hand palm rule, or Maxwell cork screw rule or right-hand thumb rule.
Question 82 Marks
Current of 0.5 A is flowing through a circular loop of radius 5 cm . Calculate the magnetic field at the center of the loop.
Answer
View full question & answer→Given :
$\quad r=5 cm=5 \times 10^{-2} m$ $ I=0.5 A $
Magnetic field at the center of the loop $ B=\frac{\mu_0}{4 \pi}\left(\frac{2 \pi I}{r}\right) $
On putting the values,
$B=\frac{4 \pi \times 10^{-7}}{4 \pi}\left(\frac{2 \pi \times 0.5}{5 \times 10^{-2}}\right)$
$=10^{-7} \times 2 \times 3.14 \times 10$
$=6.28 \times 10^{-6}$ Tesla Ans.
$\quad r=5 cm=5 \times 10^{-2} m$ $ I=0.5 A $
Magnetic field at the center of the loop $ B=\frac{\mu_0}{4 \pi}\left(\frac{2 \pi I}{r}\right) $
On putting the values,
$B=\frac{4 \pi \times 10^{-7}}{4 \pi}\left(\frac{2 \pi \times 0.5}{5 \times 10^{-2}}\right)$
$=10^{-7} \times 2 \times 3.14 \times 10$
$=6.28 \times 10^{-6}$ Tesla Ans.
Question 92 Marks
Ammeter and milli-ammeter both are made from galvanometer. Which of them has greater resistance?
Answer
View full question & answer→We know that : $S =\frac{ I _{ g } G }{ I - I _{ g }}$
It is clear from the above expression that to convert galvanometer into milli-ammeter, a shunt S of greater resistance is required than shunt resistance of ammeter.
$\because$ Resistance of ammeter, $R=\frac{G . S}{G+S}$
Hence, resistance of milli-ammeter will be more.
It is clear from the above expression that to convert galvanometer into milli-ammeter, a shunt S of greater resistance is required than shunt resistance of ammeter.
$\because$ Resistance of ammeter, $R=\frac{G . S}{G+S}$
Hence, resistance of milli-ammeter will be more.
Question 102 Marks
You are given a low resistance $R_1$, a high resistance $R_2$ and a galvanometer. How will you make such instruments from them which can measure (i) current, (ii) potential difference?
Answer
View full question & answer→(i) To measure current, low resistance $R_1$ should be joined in parallel with the galvanometer.
(ii) To measure potential difference, high resistance $R _2$ should be joined in series with the galvanometer.
(ii) To measure potential difference, high resistance $R _2$ should be joined in series with the galvanometer.
Question 112 Marks
What is being done to change the range of voltmeter?
Answer
View full question & answer→$R _{ H }=\left(\frac{ V }{i_g}\right)- G$.
On joining this high resistance to the given galvanometer in series, it will become voltmeter of range V . When high resistance $R _{ H }$ is joined with galvanometer in series, the effective resistance $R _{ V }=\left( R _{ H }+ G \right)$ of galvanometer becomes very high (camparison to G). So, when such a voltmeter is connected in parallel in a circuit to measure potential difference, then due to its high resistance, it allows very small fraction of total current to pass through it and hence gives reading of potential difference corresponding to total current flowing through the circuit.
On joining this high resistance to the given galvanometer in series, it will become voltmeter of range V . When high resistance $R _{ H }$ is joined with galvanometer in series, the effective resistance $R _{ V }=\left( R _{ H }+ G \right)$ of galvanometer becomes very high (camparison to G). So, when such a voltmeter is connected in parallel in a circuit to measure potential difference, then due to its high resistance, it allows very small fraction of total current to pass through it and hence gives reading of potential difference corresponding to total current flowing through the circuit.
Question 122 Marks
A long wire is bent as shown in the figure. Find the magnitude and direction of magnetic field at center $O$ of the circular part when a current $i$ ampere flows through it.
Answer
View full question & answer→Magnetic field at center O due to straight part of the wire
$B _1=\frac{\mu_0}{2 \pi} \times \frac{i}{ R }$
(perpendicular to the plane of the paper outward)
Magnetic field at center O due to circular loop
$B _2=\frac{\mu_0 i}{2 R }$
(perpendicular to the plane of paper outwards)
$\therefore$ Resultant magnetic field at the center
$B = B _1+ B _2$
$=\frac{\mu_0 \dot{j}}{2 \pi R }+\frac{\mu_0{ }^j}{2 R }$
$B=\frac{\mu_0 i}{2 R}\left(\frac{1}{\pi}+1\right)$
$B _1=\frac{\mu_0}{2 \pi} \times \frac{i}{ R }$
(perpendicular to the plane of the paper outward)
Magnetic field at center O due to circular loop
$B _2=\frac{\mu_0 i}{2 R }$
(perpendicular to the plane of paper outwards)
$\therefore$ Resultant magnetic field at the center
$B = B _1+ B _2$
$=\frac{\mu_0 \dot{j}}{2 \pi R }+\frac{\mu_0{ }^j}{2 R }$
$B=\frac{\mu_0 i}{2 R}\left(\frac{1}{\pi}+1\right)$
Question 132 Marks
Define the international definition of ampere.
Answer
View full question & answer→Definition of ampere : Magnetic force acting between two long, straight and parallel current carrying conductors per unit length,
If $i_1=i_2=1 A$ and $d=1 m$, then
$\frac{\delta F }{\delta l}=\frac{\mu_{ o } i_1 i_2}{2 \pi d}$
$\frac{\delta F }{\delta l}=\frac{\mu_0(1 A)(1 A)}{2 \pi(1 m)}=\frac{4 \pi \times 10^{-7}}{2 \pi}$
$\frac{\delta F }{\delta l}=2 \times 10^{-7} N / m$
1 ampere electric current is that electric current which produces a force of $2 \times 10^{-7} N / m$ on two straight, parallel, current carrying wires when it flows through them and the wires are kept at a distance of 1 m in air or vacuum on the length 1 m of the wire.
If $i_1=i_2=1 A$ and $d=1 m$, then
$\frac{\delta F }{\delta l}=\frac{\mu_{ o } i_1 i_2}{2 \pi d}$
$\frac{\delta F }{\delta l}=\frac{\mu_0(1 A)(1 A)}{2 \pi(1 m)}=\frac{4 \pi \times 10^{-7}}{2 \pi}$
$\frac{\delta F }{\delta l}=2 \times 10^{-7} N / m$
1 ampere electric current is that electric current which produces a force of $2 \times 10^{-7} N / m$ on two straight, parallel, current carrying wires when it flows through them and the wires are kept at a distance of 1 m in air or vacuum on the length 1 m of the wire.
Question 142 Marks
State conditions when resultant magnetic field at a point due to two long straight perpendicular current carrying wires can be zero.
Answer
View full question & answer→If coordinates of a point P located in $x-y$ plane is $(x, y)$, then magnetic fields due to mutually perpendicular lines is
$B _1=\frac{\mu_{ o } i_1}{2 \pi x}$
and $B _2=\frac{\mu_{ o } i_2}{2 \pi y}$
and their directions will be along $z$-axis
If there fields are equal and opposite, then their resultant field zero.
$B_1-B_2=0$
$\Rightarrow \quad B _1= B _2$
$\frac{\mu_0 i_1}{2 \pi x}=\frac{\mu_0 i_2}{2 \pi y}$
$\Rightarrow \quad \frac{i_1}{x}=\frac{i_2}{y}$
$\Rightarrow \quad \frac{i_1}{i_2}=\frac{x}{y}$
whereas these currents point outward from the origin.
$B _1=\frac{\mu_{ o } i_1}{2 \pi x}$
and $B _2=\frac{\mu_{ o } i_2}{2 \pi y}$
and their directions will be along $z$-axis
If there fields are equal and opposite, then their resultant field zero.
$B_1-B_2=0$
$\Rightarrow \quad B _1= B _2$
$\frac{\mu_0 i_1}{2 \pi x}=\frac{\mu_0 i_2}{2 \pi y}$
$\Rightarrow \quad \frac{i_1}{x}=\frac{i_2}{y}$
$\Rightarrow \quad \frac{i_1}{i_2}=\frac{x}{y}$
whereas these currents point outward from the origin.
Question 152 Marks
What is the importance of radial magnetic field in a moving coil galvanometer?
Answer
View full question & answer→Torque, $\tau= NIAB \sin \theta$
When a coil is suspended in a radial magnetic field, then in each position of coil, its plane is along any force line.
Hence, $\theta=90^{\circ}$
Therefore,$\sin \theta=1$
$\tau= NIAB$ or $\tau \propto I$
Hence, torque on a coil is proportional to the current flowing through it. So, in this way we can make scale of galvanometer linear.
When a coil is suspended in a radial magnetic field, then in each position of coil, its plane is along any force line.
Hence, $\theta=90^{\circ}$
Therefore,$\sin \theta=1$
$\tau= NIAB$ or $\tau \propto I$
Hence, torque on a coil is proportional to the current flowing through it. So, in this way we can make scale of galvanometer linear.
Question 162 Marks
A beam of electrons enters crossed region of electric field of intensity $E$ and magnetic field of intensity B. For what speed of electrons, the electron beam will remain unaffected?
Answer
View full question & answer→Electron beam will remain unaffected when the electric and magnetic force acting on them is equal in magnitude but opposite in direction.
$F _e= F _m$ or E. $e=e v B$ or $E =v B$
$\therefore v=\frac{E}{B}$
$F _e= F _m$ or E. $e=e v B$ or $E =v B$
$\therefore v=\frac{E}{B}$
Question 172 Marks
A beam of alpha particles and protons enters a uniform magnetic field with same speed. Calculate the ratio between their radii of circular path in magnetic field.
Answer
View full question & answer→Radius of circular path of any charged particle in a magnetic field
$r=\frac{m v}{q B }$
For $\alpha$-particle, $\quad r_\alpha=\frac{m_\alpha v}{2 e . B }$
For proton, $\quad r_{ p }=\frac{m_p v}{e \cdot B}$
$\therefore \quad \frac{r_\alpha}{r_p}=\frac{m_\alpha}{2 m_{ p }}=\frac{4 m_p}{2 m_p}$ $\because m \alpha=4 m_p$
or $\frac{r_\alpha}{r_p}=2 $
$\therefore \quad r_\alpha: r_p=2: 1$ Ans.
$r=\frac{m v}{q B }$
For $\alpha$-particle, $\quad r_\alpha=\frac{m_\alpha v}{2 e . B }$
For proton, $\quad r_{ p }=\frac{m_p v}{e \cdot B}$
$\therefore \quad \frac{r_\alpha}{r_p}=\frac{m_\alpha}{2 m_{ p }}=\frac{4 m_p}{2 m_p}$ $\because m \alpha=4 m_p$
or $\frac{r_\alpha}{r_p}=2 $
$\therefore \quad r_\alpha: r_p=2: 1$ Ans.
Question 182 Marks
There is no change in the energy of a moving charged particle in a plane magnetic field. Explain, why?
Answer
View full question & answer→Force exerted on a moving charged particle in magnetic field is perpendicular to the velocity of the particle. Hence, no work is done by the magnetic field on the particle. Hence the velocity of the particle will remain constant and so its kinetic energy will be unchanged.
Question 192 Marks
Write an expression $F$ in vector form for a charge $q$ moving with a velocity $\vec{v}$ in a magnetic field $\vec{B}$. Obtain conditions with the help of this expression when this force is (i) maximum and (ii) minimum.
Answer
View full question & answer→Force exerted on a charge $q$ moving with a velocity $\vec{v}$ in a magnetic field $\vec{B}$
$\overrightarrow{ F }=q(\vec{v} \times \overrightarrow{ B })$
(i) $\quad|\vec{v} \times \overrightarrow{ B }|=v B \sin \theta$
when $\theta=90^{\circ}, \vec{v} \times \overrightarrow{ B }$ then $\sin \theta=1$ which is the maximum value of $\sin \theta$, i.e, when $\vec{v} \times \overrightarrow{ B }$ then $|\vec{v} \times \overrightarrow{ B }|$ will be maximum. Hence magnitude of $\vec{F}$ will be maximum
(ii) When $\theta=0^{\circ}$. i.e. $\vec{v} \| \overrightarrow{ B }$ then $\sin \theta=0$
hence $\quad|\vec{v} \times \overrightarrow{ B }|=0$
or $\quad \vec{F}=0$.
i.e. when $\vec{v} \| \vec{d}$ then
magnitude of $\overrightarrow{ F }$ will be minimum.
$\overrightarrow{ F }=q(\vec{v} \times \overrightarrow{ B })$
(i) $\quad|\vec{v} \times \overrightarrow{ B }|=v B \sin \theta$
when $\theta=90^{\circ}, \vec{v} \times \overrightarrow{ B }$ then $\sin \theta=1$ which is the maximum value of $\sin \theta$, i.e, when $\vec{v} \times \overrightarrow{ B }$ then $|\vec{v} \times \overrightarrow{ B }|$ will be maximum. Hence magnitude of $\vec{F}$ will be maximum
(ii) When $\theta=0^{\circ}$. i.e. $\vec{v} \| \overrightarrow{ B }$ then $\sin \theta=0$
hence $\quad|\vec{v} \times \overrightarrow{ B }|=0$
or $\quad \vec{F}=0$.
i.e. when $\vec{v} \| \vec{d}$ then
magnitude of $\overrightarrow{ F }$ will be minimum.
Question 202 Marks
If current flowing in a circular coil is doubled and radius is halved, then what would be the change in the produced magnetic field at the center of the coil?
Answer
View full question & answer→Magnetic field produced at the center of the circular coil
$ B=\frac{\mu_0 NI}{2 R} $
Now if, $I ^{\prime}=2 I$ and radius $r^{\prime}=r / 2$ then magnetic field
$ \begin{array}{ll} & B^{\prime}=\frac{\mu_0 NI^{\prime}}{2 R^{\prime}}=\frac{\mu_0 N(2 I)}{2(R / 2)} \\
\therefore & B^{\prime}=4 B \end{array} $
Magnetic field will become four times.
$ B=\frac{\mu_0 NI}{2 R} $
Now if, $I ^{\prime}=2 I$ and radius $r^{\prime}=r / 2$ then magnetic field
$ \begin{array}{ll} & B^{\prime}=\frac{\mu_0 NI^{\prime}}{2 R^{\prime}}=\frac{\mu_0 N(2 I)}{2(R / 2)} \\
\therefore & B^{\prime}=4 B \end{array} $
Magnetic field will become four times.
Question 212 Marks
An electron is not deflected while passing through any field. Is it possible that there is no magnetic field present at that place? Explain.
Answer
View full question & answer→Exerting force in magnetic field on a moving electron with velocity $\vec{v}$
$\overrightarrow{ F }=-e(\vec{v} \times \overrightarrow{ B })$
or $\overrightarrow{ F }=-e v B \sin \theta$
If $\theta=0^{\circ}$ or $180^{\circ}$, the magnitude of force F will be zero. Hence, an electron will not be deflected if it moves parallel or opposite to the direction. So, we cannot say that magnetic field is absent.
$\overrightarrow{ F }=-e(\vec{v} \times \overrightarrow{ B })$
or $\overrightarrow{ F }=-e v B \sin \theta$
If $\theta=0^{\circ}$ or $180^{\circ}$, the magnitude of force F will be zero. Hence, an electron will not be deflected if it moves parallel or opposite to the direction. So, we cannot say that magnetic field is absent.
Question 222 Marks
Under what condition, force on moving charged particle in a magnetic field will be (i) maximum (ii) minimum?
Answer
View full question & answer→From, $F =q \nu$ B $\sin \theta$
(i) when $\theta=90^{\circ}$, then $F =q v B$ which is the maximum value.
(ii) when $\theta=0^{\circ}$ or $\theta=180^{\circ}$ then $F =0$, which is the minimum value.
When a charged particle moves perpendicular to the direction of magnetic field then maximum force is exerted on it and when it moves in the direction of magnetic field or opposite to the direction of magnetic field, then the force everted on it is minimum.
(i) when $\theta=90^{\circ}$, then $F =q v B$ which is the maximum value.
(ii) when $\theta=0^{\circ}$ or $\theta=180^{\circ}$ then $F =0$, which is the minimum value.
When a charged particle moves perpendicular to the direction of magnetic field then maximum force is exerted on it and when it moves in the direction of magnetic field or opposite to the direction of magnetic field, then the force everted on it is minimum.
Question 232 Marks
On changing a current carrying coil of N turns and R radius into a straight wire, how many times the magnetic field at distance R will be than at the magnitude at the center of the coil?
Answer
View full question & answer→Magnetic field at the center of coil
$ B_0=\frac{\mu_0 NI}{2 R} $
and magnetic field due to long straight wire
$ B=\frac{\mu_0 I}{2 \pi R} $
Dividing equation (1) by (2)
$ \frac{B_0}{B}=\frac{\frac{\mu_0 N I}{2 R}}{\frac{\mu_0 I}{2 \pi R}}=N \pi $
$ B_0=\frac{\mu_0 NI}{2 R} $
and magnetic field due to long straight wire
$ B=\frac{\mu_0 I}{2 \pi R} $
Dividing equation (1) by (2)
$ \frac{B_0}{B}=\frac{\frac{\mu_0 N I}{2 R}}{\frac{\mu_0 I}{2 \pi R}}=N \pi $
Question 242 Marks
Unit of magnetic flux is weber. State its equivalent electric unit
Answer
View full question & answer→$\because \quad \phi=$ B.A
$=\frac{\text { Newton }}{\text { Coulomb } \times \frac{\text { metre }}{\text { second }}} \times$ metre $^2$
$=\frac{\text { Newton } \times \text { metre }}{\text { Coulomb }} \times$ second
$=\frac{\text { Joule }}{\text { Coulomb }} \times$ second
$=$ Volt $\times$ second
$=\frac{\text { Newton }}{\text { Coulomb } \times \frac{\text { metre }}{\text { second }}} \times$ metre $^2$
$=\frac{\text { Newton } \times \text { metre }}{\text { Coulomb }} \times$ second
$=\frac{\text { Joule }}{\text { Coulomb }} \times$ second
$=$ Volt $\times$ second
Question 252 Marks
Write conclusion of Oersted's experiment.
Answer
View full question & answer→A current carrying conductor produces magnetic field around it whose magnitude and direction depends on the magnitude and direction of current flowing in the conductor.