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 $=MB(\cos {\theta }_1-\cos {\theta }_2)$

(i) ${\theta }_1={60}^{\circ },{\theta }_2={90}^{\circ }$

$$\begin{array}{rlrl}\therefore \text{ Work done }& =MB(\cos {60}^{\circ }-\cos {90}^{\circ })& =MB(\frac{1}{2}-0)=\frac{1}{2}MB1/2& =\frac{1}{2}\times 6\times 0\cdot 44\mathrm{J}=1.32\mathrm{J}1/2\end{array}$$

(ii) ${\theta }_1={60}^{\circ },{\theta }_2={180}^{\circ }$

$\therefore$ Work done $=MB(\cos {60}^{\circ }-\cos {180}^{\circ })$

$$\begin{array}{rlr}& =MB(\frac{1}{2}-(-1))=\frac{3}{2}MB1/2& =\frac{3}{2}\times 6\times 0.44\mathrm{J}=3.96\mathrm{J}1/2\end{array}$$

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

(b)

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

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, $\overrightarrow{B}=0.44\mathrm{T}$

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

(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

$$\begin{array}{}\text{(1)}& W_1& =-MB(\cos {\theta }_2-\cos {\theta }_1)& =-6\times 0.44(\cos {90}^{\circ }-\cos {60}^{\circ })=-2.64\times (-\frac{1}{2})W_1& =1.32\text{ Joule. }\end{array}$$

(ii) Work done in aligning the magnet opposite to magnetic field. i.e. ${\theta }_2={180}^{\circ },{\theta }_1={60}^{\circ }$

$$\begin{array}{rlrl}{W}_2& =-MB(\cos {180}^{\circ }-\cos {60}^{\circ })=-6\times 0.44[-1-\frac{1}{2}]& =-2.64\times (-\frac{3}{2})=+3.96\text{ Joule }{W}_2& =+3.96\text{ Joule }\end{array}$$

(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,

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

(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,

$$\tau =MB\mathrm{Sin}\theta$$

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

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