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(a) A circular coil of 30 turns and radius $8.0 \mathrm{~cm}$ carrying a current of $6.0 \mathrm{~A}$ is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in
(a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
(b) Would your answer change, if the circular coil in
(a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Solution:
1802 Upvotes
Verified Answer
(a) Given, $\mathrm{N}=30, \mathrm{r}=8 \mathrm{~cm}=0.08 \mathrm{~m}$
$$
\begin{aligned}
&\mathrm{A}=\pi(0.08)^2 \mathrm{~m}^2, \mathrm{I}=6 \mathrm{~A}, \mathrm{~B}=1 \mathrm{~T}, \theta=60^{\circ} \\
&\Rightarrow \sin \theta=0.866
\end{aligned}
$$
Restoring torque $=$ deflecting torque
$=\mathrm{NIAB} \sin \theta$
$$
\tau=30 \times 6 \times \pi \times 0.0064 \times 1 \times 0.866
$$
$$
=3.1341 \mathrm{Nm} \text {. }
$$
(b) The torque depends upon area of coil and not shape. Answer will be same. No change.
$$
\begin{aligned}
&\mathrm{A}=\pi(0.08)^2 \mathrm{~m}^2, \mathrm{I}=6 \mathrm{~A}, \mathrm{~B}=1 \mathrm{~T}, \theta=60^{\circ} \\
&\Rightarrow \sin \theta=0.866
\end{aligned}
$$
Restoring torque $=$ deflecting torque
$=\mathrm{NIAB} \sin \theta$
$$
\tau=30 \times 6 \times \pi \times 0.0064 \times 1 \times 0.866
$$
$$
=3.1341 \mathrm{Nm} \text {. }
$$
(b) The torque depends upon area of coil and not shape. Answer will be same. No change.
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