Q142 / 170 · Manipal (MET) 2015 · Physics
A circular coil of 100 turns has an effective radius of 0.05 m and carries a current of 0.1 A . How much work is required to turn it in an external magnetic field of $1.5 \mathrm{~Wb} / \mathrm{m}^2$ through $180^{\circ}$ about its axis perpendicular to the magnetic field? The plane of the coil is initially perpendicular to the magnetic field.
- A0.456 J
- B2.65 J
- Correct answerC0.2355 J
- D3.95 J
Solution
The potential energy of circular loop in the magnetic field $=-M B \cos \theta$, where $\theta$ is the angle between normal to plane of coil and magnetic field $B$.
Initial potential energy,
$U_i=-N i A B \cos 0^{\circ}=-N i A B$
When coil is turned through $180^{\circ}$, therefore final potential energy, $U_f=-N i A B \cos 180^{\circ}=N i A B$ $\therefore$ Required work, $W=$ gain in potential energy $=U_f-U_i=N i A B-(-N i A B)=2NiAB$
Here, $N=100, \quad i=0.1 \mathrm{~A}, \quad B=1.5 \mathrm{~Wb} / \mathrm{m}^2$ and radius $r=0.05 \mathrm{~m}$ $\therefore \quad A=\pi r^2=3.14 \times(0.05)^2=7.85 \times 10^{-3} \mathrm{~m}^2$ $\therefore$ Work, $\quad W=2 \times 100 \times 0.1 \times 7.85 \times 10^{-3} \times 1.5$ $\Rightarrow \quad W=0.2355 \mathrm{~J}$
When coil is turned through $180^{\circ}$, therefore final potential energy, $U_f=-N i A B \cos 180^{\circ}=N i A B$ $\therefore$ Required work, $W=$ gain in potential energy $=U_f-U_i=N i A B-(-N i A B)=2NiAB$
Here, $N=100, \quad i=0.1 \mathrm{~A}, \quad B=1.5 \mathrm{~Wb} / \mathrm{m}^2$ and radius $r=0.05 \mathrm{~m}$ $\therefore \quad A=\pi r^2=3.14 \times(0.05)^2=7.85 \times 10^{-3} \mathrm{~m}^2$ $\therefore$ Work, $\quad W=2 \times 100 \times 0.1 \times 7.85 \times 10^{-3} \times 1.5$ $\Rightarrow \quad W=0.2355 \mathrm{~J}$
Hence, the correct answer is option: 'C'.