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,

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

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

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