3 Marks Question

Question 13 Marks

Magnetic moment of a small bar magnet is 0.24 newton metre tesla ${ }^{-1}$. Calculate magnetising field of the magnet at a distance of 10 cm from the centre of the magnet on its axis.

Answer

Given : $m=0.24$ newton metre tesla ${ }^{-1}$
$ r=10 cm=0.10 \text { metre } $
$\therefore$ In axial position, $ B=\frac{\mu_0}{4 \pi}\left(\frac{2 m}{r^3}\right) $
On putting the values, $B =\frac{4 \pi \times 10^{-7}}{4 \pi}\left(\frac{2 m}{r^3}\right)$
$=10^{-7}\left(\frac{2 m}{r^3}\right)=\frac{10^{-7}(2 \times 0.24)}{(0.10)^3}$
$=4.8 \times 10^{-5}$ tesla $\quad$ Ans.

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Question 23 Marks

0.5 A of current is flowing through a coil of 25 turns and area $200 cm^2$. What is the pole strength of 10 cm length of bar magnet whose magnetic moment is equal to magnetic moment of the coil?

Answer

Magnetic moment of bar magnet $=$ Magnetic moment of the coil
$ \therefore \quad q_m \times 2 l=\text { NIA } $
Here, pole strength $ \left(q_m\right)=\frac{\text { NIA }}{2 l} $
On putting the values,
$ \begin{aligned} q_m & =\frac{25 \times 0.5 \times\left(200 \times 10^{-4}\right)}{10 \times 10^{-2}} A-m \\ & =2.5 A-m . \end{aligned} $

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Question 33 Marks

A coil of area $5 cm^2$ is kept in a uniform magnetising field of $2.5 N / Am$. Number of turns in the coil is 100 . If 0.2 A of current is flowing, then calculate:
(a) Magnetising dipole moment,
(b) Maximum moment of force.

Answer

Given : Area of the bottom of the coil $A =5 cm^2=$ $5 \times 10^{-4} m^2$, Magnetic field intensity B $=2.5 N / Am$, Number of turns, $N =100$, Current flowing in the coil $I =0.2 A$
(a) Magnetising dipole moment $m=$ NIA $ \begin{aligned} m & =100 \times 0.2 A \times 5 \times 10^{-4} m^2 \\ & =0.01 A-m^2 \end{aligned} $
(b) Magnetic moment of force $ \begin{aligned} \tau_{\max } & =m B \\ & =0.01 A-m^2 \times 2.5 N / A m \\ & =0.025 N-m \end{aligned} $

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Question 43 Marks

Write relation between $B , H , I , \chi$ and $\mu$.

Answer

Magnetic suspectibility : In a small magnetised field, intensity of magnetisation (I) of the substance is proportional to magnetic field $( H )$,
which means $I \propto H$ or, $ \begin{array}{c} I=\chi H \\or \chi=\frac{I}{H} \end{array} $
Here, is a constant which is known as magnetic susceptibility of the substance.
Magnetic permeability : Capacity of passing of magnetic field lines from any medium is known as magnetic permeability of that medium. Mathematically, it is the ratio of magnetic induction induced B inside the substance and magnetic field H . It is represented by $\mu$. It is a scalar quantity.
$ \mu=\frac{B}{H} $ On dividing equation (1) by (2), $ \frac{\mu}{\chi}=\frac{\frac{B}{H}}{\frac{I}{H}} $

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Question 53 Marks

What is magnetic flux density? Obtain a relation between magnetic flux and magnetic induction.

Answer

Number of magnetic field lines passing through surface of unit area placed perpendicular to the force lines is called magnetic flux density. It is a vector quantity which is denoted by $\overrightarrow{ B }$. It is also called magnetisation field or intensity of magnetisation field or magnetic induction. Its unit is weber/ meter $^2$ or tesla or newton/ampere-metre.
If in the magnetic field of intensity $\vec{B}$, a surface of infinitely small surface is located at any point, then magnetic flux associated with it will be $\delta \phi_{ B ^{\prime}}$
$ \delta \phi_{B}=\overrightarrow{B} \cdot \overrightarrow{\delta S}=B \delta s \cos \theta $
And total associated magnetic flux $\phi_{ B }=\int_{ S } \overrightarrow{ B } . \overrightarrow{\delta S }$
Magnetic flux is a scalar quantity but magnetic flux density is a vector quantity.

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Question 63 Marks

Define a magnet and write its characteristics of directive property and pole strength.

Answer

Such a substance which attracts other magnetic substances such as iron etc. towards itself and when it is suspended freely, then it rests in north-south direction. Its two and equal opposite poles are separated by some distance.
Directive property : When any bar magnet is suspended freely, it always rests in a definite direction. Its one end is towards the north and the other is towards the south which are known as north pole and south pole respectively. A bar magnet always exhibits north and south direction.
Pole strength : Strength of magnetisation of magnetic poles is known as pole strength. Pole strength of both poles of a bar magnet is same. If a bar magnet is cut perpendicular to its axis, then pole strength of poles of each part will almost be same as of original magnet and if it is cut in to equal parts along its axis, then pole strength of each part will be half that of original bar magnet.

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Question 73 Marks

A short bar magnet has a magnetic moment of $0.48 JT ^{-1}$.Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Answer

Given : $m=0.48 JT ^{-1}, r=10 cm=10 \times 10^{-2} m=$ $10^{-1} m, B =$ ?
(i) Along the axial line,
$B=\frac{\mu_0 m}{2 \pi r^3}$
On putting the values,
$B =\frac{10^{-7} \times 0.48}{2 \times 3.14 \times\left(10^{-1}\right)^3}$
$=\frac{48 \times 10^{-4}}{628}$
$=0.96 \times 10^{-4} T$
$=0.96 G$ Along the $N - S$ direction
(ii) Along the equatorial line,
$B=\frac{\mu_0 m}{4 \pi r^3}$
$=\frac{1}{2}\left|\frac{\mu_0 m}{2 \pi r^3}\right|=\frac{1}{2}(0.96)$
$=0.48 G$ Along the N-S direction

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Question 83 Marks

A closely wound solenoid of 2000 turns and area of cross-section$1.6 \times 10^{-4} m^2$, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of$7.5 \times 10^{-2} T$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?

Answer

Given : Number of turns in solenoid N = 2000 Area of corss-section of solenoid $A =1.6 \times 10^{-4} m^2$ Current carried by solenoid $I =4.0 A, \theta=30^{\circ}$,Uniform horizontal magnetic field B = $7.5 \times 10^{-2} T$
(a) Magnetic moment m = NIA
$=2000 \times 4.0 \times 1.6 \times 10^{-4}$
$=1.28 Am ^2$
It depends upon the direction of magnetic field and direction can be obtained with the help of right hand screw rule.
(b) Since field is uniform, force is zero.
Torque, $\quad \tau=$ NBIA $\sin \theta$
On putting values, $\quad \tau=2000 \times 7.5 \times 10^{-2} \times 4.0$$\times 1.6 \times 10^{-4} \times \sin 30^{\circ}$
$\begin{array}{c}=150 \times 32 \times 10^{-3} \\
=4800 \times 10^{-5}=0.048 Nm \end{array}$
Its direction is such that it tries to bring axis of the solenoid (that is, moment vector) along the direction of $\overrightarrow{ B }$.

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Question 93 Marks

A bar magnet of magnetic moment $1.5 JT ^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in the cases (i) and (ii)?

Answer

Given: Magnetic moment m =$m=1.5 JT ^{-1}$, Uniform magnetic field B = 0.22 T
(a) (i) Here $\theta_1=0^{\circ}$ (along magnetic field) and $\theta_2=90^{\circ}$(perpendicular to magnetic field)
$\because W =-m B\left(\cos \theta_2-\cos \theta_1\right)$
$=-1.5 \times 0.22 \times\left(\cos 90^{\circ}-\cos 0^{\circ}\right)$
$=-0.33(0-1)$
$=0.33 J$
(ii) Here, $\quad \theta_1=0^{\circ}, \theta_2=180^{\circ}$
$\therefore \quad W =-1.5 \times 0.22 \times\left(\cos 180^{\circ}-\cos 0^{\circ}\right)$
$=-0.33 \times(-1-1)$
$=-0.33 \times(-2)$
$=0.66 J$
(b) Torque $\quad \tau=m B \sin \theta$
(i)$\theta=90^{\circ}$
$\tau=1.5 \times 0.22 \times \sin 90^{\circ}$
$=0.33 J$
Torque of magnitude which tends to bring the magnetic moment vector A along B .
(ii) Here given : $\theta=180^{\circ}$
$\begin{array}{l}\tau=1.5 \times 0.22 \times \sin 180^{\circ} \\ \tau=0 \quad\left(\text { Since } \sin 180^{\circ}=0\right)\end{array}$

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Question 103 Marks

A short bar magnet of magnetic moment m =$=0.32 JT ^{-1}$ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable equilibrium and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Answer

Given: Magnetic moment of bar magnet m = 0.32 JT^-1,unifom magnetic field B = 0.15 T
(a) For stable equilibrium, magnetic moment M of the magnet must be parallel to the magnetic field B, which means $\theta=0^{\circ}$
Potential energy in this condition
$U =-\overrightarrow{ m } \cdot \overrightarrow{ B }=-m B \cos \theta$
On putting values, $U =-m B \cos 0$
$U =-m B \times 1 \quad \because \cos 0=1$
$U =-0.048 J=-4.8 \times 10^{-2} J$
(b) For unstable equilibrium, $\theta=\pi$ which means $\overrightarrow{ m }$ and $\vec{B}$ are relatively opposite.
Potential energy in this condition
$U =-\overrightarrow{ m } \cdot \overrightarrow{ B }=-m B \cos 180^{\circ}$
$U =+m B \quad \because \cos 180^{\circ}=-1$
On putting values, $U =+(0.32) \times(0.15)$
$=+4.8 \times 10^{-2} J$

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