Answer

Magnetic field strength, $B=0.25 \mathrm{~T}$

Torque on the bar magnet, $T=4.5 \times 10^{-2} \mathrm{~J}$

Angle between the bar magnet and the external magnetic field, $\theta=30^{\circ}$

Torque is related to magnetic moment $(M)$ as:

$T=M B \sin \theta$

$\therefore M=\frac{T}{B \sin \theta}$

$$ =\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 \mathrm{~J} \mathrm{~T}^{-1} $$

Hence, the magnetic moment of the magnet is $0.36 \mathrm{~J} \mathrm{~T}^{-1}$.

Answer

Moment of the bar magnet, $M=0.32 \mathrm{~J} \mathrm{~T}^{-1}$

External magnetic field, $B=0.15 \mathrm{~T}$

(a)The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle $\theta$, between the bar magnet and the magnetic field is $0^{\circ}$.

Potential energy of the system $=-M B \cos \theta$

$=-0.32 \times 0.15 \cos 0^{\circ}$

$=-4.8 \times 10^{-2} \mathrm{~J}$

(b)The bar magnet is oriented $180^{\circ}$ to the magnetic field. Hence, it is in unstable equilibrium.

$\theta=180^{\circ}$

Potential energy $=-M B \cos \theta$

$=-0.32 \times 0.15 \cos 180^{\circ}$

$=4.8 \times 10^{-2} \mathrm{~J}$

Answer

Number of turns in the solenoid, $n=800$

Area of cross-section, $A=2.5 \times 10^{-4} \mathrm{~m}^{2}$

Current in the solenoid, $I=3.0 \mathrm{~A}$

A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

$M=n I A$ $=800 \times 3 \times 2.5 \times 10^{-4}$

$=0.6 \mathrm{~J} \mathrm{~T}^{-1}$

Answer

Magnetic field strength, $B=0.25 \mathrm{~T}$

Magnetic moment, $M=0.6 \mathrm{~T}^{-1}$

The angle $\theta$, between the axis of the solenoid and the direction of the applied field is $30^{\circ}$.

Therefore, the torque acting on the solenoid is given as:

$$ \begin{aligned} \tau & =M B \sin \theta \\ & =0.6 \times 0.25 \sin 30^{\circ} \\ & =7.5 \times 10^{-2} \mathrm{~J} \end{aligned} $$

Answer

(a)Magnetic moment, $M=1.5 \mathrm{~J} \mathrm{~T}^{-1}$

Magnetic field strength, $B=0.22 \mathrm{~T}$

(i)Initial angle between the axis and the magnetic field, $\theta_{1}=0^{\circ}$

Final angle between the axis and the magnetic field, $\theta_{2}=90^{\circ}$

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

$$ \begin{aligned} W & =-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \\ & =-1.5 \times 0.22\left(\cos 90^{\circ}-\cos 0^{\circ}\right) \\ & =-0.33(0-1) \\ & =0.33 \mathrm{~J} \end{aligned} $$

(ii) Initial angle between the axis and the magnetic field, $\theta_{1}=0^{\circ}$

Final angle between the axis and the magnetic field, $\theta_{2}=180^{\circ}$

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

$$ \begin{aligned} W & =-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \\ & =-1.5 \times 0.22\left(\cos 180-\cos 0^{\circ}\right) \\ & =-0.33(-1-1) \\ & =0.66 \mathrm{~J} \end{aligned} $$

(b) For case (i): $\theta=\theta_{2}=90^{\circ}$

$\therefore$ Torque, $\tau=M B \sin \theta$

$=1.5 \times 0.22 \sin 90^{\circ}$

$=0.33 \mathrm{~J}$

$\underline{\text { For case (ii): }} \theta=\theta_{2}=180^{\circ}$

$\therefore$ Torque, $\tau=M B \sin \theta$

$=M B \sin 180^{\circ}=0 \mathrm{~J}$

Answer

Number of turns on the solenoid, $n=2000$

Area of cross-section of the solenoid, $A=1.6 \times 10^{-4} \mathrm{~m}^{2}$

Current in the solenoid, $I=4 \mathrm{~A}$

(a)The magnetic moment along the axis of the solenoid is calculated as:

$M=n A I$

$=2000 \times 1.6 \times 10^{-4} \times 4$

$=1.28 \mathrm{Am}^{2}$

(b)Magnetic field, $B=7.5 \times 10^{-2} \mathrm{~T}$

Angle between the magnetic field and the axis of the solenoid, $\theta=30^{\circ}$

Torque, $\tau=M B \sin \theta$

$=1.28 \times 7.5 \times 10^{-2} \sin 30^{\circ}$

$=4.8 \times 10^{-2} \mathrm{Nm}$

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is $4.8 \times 10^{-2} \mathrm{Nm}$.

Answer

Magnetic moment of the bar magnet, $M=0.48 \mathrm{~J} \mathrm{~T}^{-1}$

Distance, $d=10 \mathrm{~cm}=0.1 \mathrm{~m}$

The magnetic field at distance $d$, from the centre of the magnet on the axis is given by the relation:

$$ B=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}} $$

Where,

$$ \begin{aligned} & \mu_{0}=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1} \\ & \therefore B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}} \\ & \quad=0.96 \times 10^{-4} \mathrm{~T}=0.96 \mathrm{G} \end{aligned} $$

The magnetic field is along the $\mathrm{S}-\mathrm{N}$ direction.

The magnetic field at a distance of $10 \mathrm{~cm}$ (i.e., $d=0.1 \mathrm{~m}$ ) on the equatorial line of the magnet is given as:

$$ \begin{aligned} B & =\frac{\mu_{0} \times M}{4 \pi \times d^{3}} \\ & =\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi(0.1)^{3}} \\ & =0.48 \mathrm{G} \end{aligned} $$

The magnetic field is along the $\mathrm{N}-\mathrm{S}$ direction.

The new null points will be located $11.1 \mathrm{~cm}$ on the normal bisector.