Question

A coil formed by wrapping $50$ turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of ${30}^o$, with the direction of the field. When the magnetic field is increased uniformly from $200\mu T$ to $600\mu T$ in $0.4s$, an emf of magnitude $80.0mV$ is induced in the coil. What is the total length of the wire in metres?

Answer 1

$e=\frac{N\Delta \varphi }{\Delta t}=\left(\frac{N\Delta B}{\Delta t}\right)=\cos {30}^o$
$\therefore S=\left(\frac{e\Delta t}{\Delta B}\right)\sec {30}^o=\left[\frac{(80\times {10}^{-3})\left(0.4\right)}{\left(50\right)(400\times {10}^{-6})}\right]\left[\frac{2}{\sqrt{3}}\right]=1.85m$
Side of square $=1.36m$
Total length of wire $=50(4\times 1.36)=272m$