Question

A square wire loop of $10\text{cm}$ side lies at right angles to a uniform magnetic field of $20\text{T}$ in vertically downwards direction. A $10\text{V}$ light bulb is in a series with the loop as shown in the fig. The magnetic field is decreasing steadily to zero over a time interval $\Delta t$. The bulb will shine with full brightness if $\Delta t$ is equal to

• A. $20\text{ms}$
• B. $0.02\text{ms}$
• C. $2\text{ms}$
• D. $0.2\text{ms}$

Answer 1

The correct option is A $20\text{ms}$

Area of the square loop is
$S=100{\text{cm}}^2={10}^{-2}{\text{m}}^2$
Taking direction of area along direction of $\overrightarrow{B},(\cos \theta =\cos 0^{\circ }=1)$
$\varphi =BS\cos 0^{\circ }=BS$
$\Rightarrow$Rate of change of flux linked with loop is,
$\frac{\Delta \varphi }{\Delta t}=S\left(\frac{\Delta B}{\Delta t}\right)...\left(i\right)$
induced emf across loop is,
$\epsilon =-\frac{\Delta \varphi }{\Delta t}...\left(ii\right)$
The bulb will shine with full brightness if induced emf in loop is equal to $10\text{V}$.
$\Rightarrow$From Eq $\left(i\right)\&\left(ii\right)$,
$10=-S\left(\frac{\Delta B}{\Delta t}\right)$
$10=-{10}^{-2}\times \left(\frac{0-20}{\Delta t}\right)$
$\Delta t=20\times {10}^{-3}\text{s}$
$\therefore \Delta t=20\text{ms}$