A circular coil of 30 turns and radius $8 \mathrm{~cm}$ carrying a current of $6 \mathrm{~A}$ is suspe-ded…
A circular coil of 30 turns and radius $8 \mathrm{~cm}$ carrying a current of $6 \mathrm{~A}$ is suspe-ded vertically in a uniform horizontal magnetic field of magnitude $1.0 \mathrm{~T}$. The field lines make an angle of $20^{\circ}$ with the normal of the coil. The magnitude of the counter torque that must be applied to prevent the coil from turning is
$5.4 \mathrm{Nm}$
$7.2 \mathrm{Nm}$
$3.6 \mathrm{Nm}$
$1.8 \mathrm{Nm}$
Solution
Number of turns, $\mathrm{N}=30$ turns
Radius, $\mathrm{r}=8 \mathrm{~cm}=0.08 \mathrm{~m}$
Current, $\mathrm{i}=6 \mathrm{~A}$
Magnetic field, $\mathrm{B}=1 \mathrm{~T}$
The magnitude of the counter tarque will be
$\begin{aligned}
& \tau=\mathrm{NiAB} \sin \theta \\
& =\mathrm{N} \text { i } \pi \mathrm{r}^2 \mathrm{~B} \sin 30^{\circ} \\
& =30 \times 6 \times 3.14 \times(0.08)^2 \times 1 \times \frac{1}{2} \\
& \tau=1.8 \mathrm{Nm}
\end{aligned}$