Question

Two infinitely long conductors carries current $I$, each along the $x$-axis and $y$-axis, respectively. A small circular coil of radius $R$, mass $m$ and having $N$ turns is also carrying current $I$ in anticlockwise direction. The plane of the loop is parallel to the $xy$-plane, with its centre at a distance $d$ along $z$-axis. Find the angular acceleration (in ${\text{rad s}}^{-2}$) of the loop.

Take: $d=\frac{\pi }{\sqrt{2}},I=10\text{A},N=1000,m=1\text{g}$

Answer 1

The magnetic field due to the infinite wire at the centre of the loop is $B=\frac{{\mu }_{o}I}{2\pi d}$.

So, net magnetic field will be,

$B_o=\frac{\sqrt{2}{\mu }_{o}I}{2\pi d}=\frac{{\mu }_{o}I}{\sqrt{2}\pi d}$

Now, torque on the loop,

$\tau =MB\sin \theta =NIAB\sin \theta$

$\Rightarrow \tau =NI\left(\pi R^2\right)\left(\frac{{\mu }_{o}I}{\sqrt{2}\pi d}\right)\sin {90}^{\circ }$

$\Rightarrow \tau =\frac{{\mu }_{o}N{I}^2{R}^2}{\sqrt{2}d}=I\alpha$

$\Rightarrow \frac{{\mu }_{o}N{I}^2{R}^2}{\sqrt{2}d}=\left(\frac{m{R}^2}{2}\right)\alpha$

$\Rightarrow \alpha =\frac{\sqrt{2}{\mu }_{o}N{I}^2}{md}=\frac{\sqrt{2}\times 4\pi \times {10}^{-7}\times 1000\times {10}^2}{0.001\times \pi /\sqrt{2}}$

$\Rightarrow \alpha =80{\text{rad/s}}^2$