Question

A closed coil having $50\text{turns}$, area $300{\text{cm}}^2$, is rotated from a position where its plane makes an angle of ${60}^{\circ }$ with a magnetic field of flux density $2.0\text{T}$ to a position perpendicular to the field in a time of $10\text{s}$. What is the average $\text{emf}$ induced in the coil?

Answer 1

Initially, the flux linked with each turn of the coil is given as,

$\varphi =\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta =BA\cos {60}^{\circ }$

Substituting the values, we get

$\varphi =2.0\times 300\times {10}^{-4}\times 0.5$

$\Rightarrow \varphi =3\times {10}^{-2}\text{Wb}$

Finally, flux linked with each turn of the coil is,

${\varphi }^{\prime }=BA\cos 0^{\circ }=BA$

$\Rightarrow {\varphi }^{\prime }=2.0\times 300\times {10}^{-4}=6.0\times {10}^{-2}\text{Wb}$

Change in flux in this process is,

$\Delta \varphi ={\varphi }^{\prime }-\varphi$

$=6.0\times {10}^{-2}-3.0\times {10}^{-2}$

$=3.0\times {10}^{-2}\text{Wb}$

This change took place in $0.1\text{s}$. The magnitude of the average emf induced in the coil is,

$E=N\frac{\Delta \varphi }{\Delta t}=50\times \frac{3.0\times {10}^{-2}}{0.1}=15\text{V}$

Accepted answers: $15,15.0,15.00$