Question: Q. 3. A bar magnet of magnetic moment $6 \mathrm{~J} / \mathrm{T}$ is aligned at $60^{\circ}$ with a uniform external magnetic field of $0.44 \mathrm{~T}$. 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). $U$ [Delhi & OD 2018]

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Solution:

Ans. (a) Formula and

Calculation of work done in the two cases $(1+1)$

(b) Calculation of torque in case (ii)

$$ 1 $$

(a) Work done $=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)$

(i) $\theta_{1}=60^{\circ}, \theta_{2}=90^{\circ}$

$$ \begin{aligned} \therefore \quad \text { Work done } & =M B\left(\cos 60^{\circ}-\cos 90^{\circ}\right) \ & =M B\left(\frac{1}{2}-0\right)=\frac{1}{2} M B \quad 1 / 2 \ & =\frac{1}{2} \times 6 \times 0 \cdot 44 \mathrm{~J}=1.32 \mathrm{~J} \quad 1 / 2 \end{aligned} $$

(ii) $\theta_{1}=60^{\circ}, \theta_{2}=180^{\circ}$

$\therefore \quad$ Work done $=M B\left(\cos 60^{\circ}-\cos 180^{\circ}\right)$

$$ \begin{aligned} & =M B\left(\frac{1}{2}-(-1)\right)=\frac{3}{2} M B \quad 1 / 2 \ & =\frac{3}{2} \times 6 \times 0.44 \mathrm{~J}=3.96 \mathrm{~J} \quad 1 / 2 \end{aligned} $$

[Also accept calculations done through changes in potential energy.]

(b)

$$ \text { Torque }=|\vec{M} \times \vec{B}|=M B \sin \theta $$

For

$$ \theta=180^{\circ} \text {, } $$

$1 / 2$

We have

$$ \text { Torque }=6 \times 0.44 \sin 180^{\circ}=0 $$

[If the student straight away writes that the torque is zero since magnetic moment and magnetic field are anti parallel in this orientation, award full marks] $1 / 2$

[CBSE Marking Scheme 2018]

Detailed Answer :

Given, $\vec{B}=0.44 \mathrm{~T}$

$\mathrm{M}=6 \mathrm{~J} / \mathrm{T}$

(a) (i) $\theta_{1}=60^{\circ}, \theta_{2}=90^{\circ}$ since magnet is placed perpendicular to magnetic field. So, work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is

$$ \begin{align*} W_{1} & =-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \ & =-6 \times 0.44\left(\cos 90^{\circ}-\cos 60^{\circ}\right)=-2.64 \times\left(-\frac{1}{2}\right) \ W_{1} & =1.32 \text { Joule. } \tag{1} \end{align*} $$

(ii) Work done in aligning the magnet opposite to magnetic field. i.e. $\theta_{2}=180^{\circ}, \theta_{1}=60^{\circ}$

$$ \begin{aligned} W_{2} & =-M B\left(\cos 180^{\circ}-\cos 60^{\circ}\right)=-6 \times 0.44\left[-1-\frac{1}{2}\right] \ & =-2.64 \times\left(-\frac{3}{2}\right)=+3.96 \text { Joule } \ W_{2} & =+3.96 \text { Joule } \end{aligned} $$

(b) The Torque on magnet aligned at angle $\theta_{2}$ is given by $\tau=$ MBsin $\theta_{2}$ in case a (ii) $\theta_{2}=180^{\circ}$ therefore

$\tau=6 \times 0.44 \times \sin 180^{\circ}=2.4 \times 0$

$\tau=0$

$\Rightarrow$ Torque in case a (ii) i.e., at $\theta_{2}=180$ position is zero. 1 AI Q. 4. A closely wound solenoid of 2000 turns and area of cross section $1.6 \times 10^{-4} \mathrm{~m}^{2}$ carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

(i) What is the magnetic moment associated with the solenoid?

(ii) What are the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{-2}$ $T$ is setup at an angle of $30^{\circ}$ with the axis of the solenoid?

U] [CBSE OD Set I, 2015]

Ans. Number of turns in solenoid,

$$ N=2000 $$

(given)

Area of cross-section of solenoid,

$$ \begin{aligned} & A=1.6 \times 10^{-4} \mathrm{~m}^{2} \quad \text { (given) } \ & I=4.0 \mathrm{~A} \quad \text { (given) } \end{aligned} $$

(i) Magnetic moment of the solenoid,

$$ m=\text { NIA } $$

$=2000 \times 1.6 \times 10^{-4} \times 4.0$

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

(ii) Net force experienced by the magnetic dipole in the uniformpmagnetic field $=0$ but it will experience torgue

Torque on a solenoid,

$$ \tau=M B \operatorname{Sin} \theta $$

$$ \vec{B}=7.5 \times 10^{-2} \mathrm{~T} \text { and } \theta=30^{\circ} \text { (given) } $$

$$ \begin{array}{rlrl} \tau & =1.28 \times 7.5 \times 10^{-2} \times \frac{1}{2} \text { (given) } \ & =4.8 \times 10^{-2} \mathrm{Nm} & \mathbf{1 1 1} / 2 \end{array} $$



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