Question

A square loop of side 4 cm is lying on a horizontal table. A uniform magnetic field of 0.5 T is directed downwards at an angle of ${60}^o$ to the vertical as shown in Fig. 25.3. If the field increases from zero to its final value in 0.2 s, the emf induced in the loop will be

• A. 1 mV
• B. 2 mV
• C. 3 mV
• D. 4 mV

Answer 1

The correct option is B

2 mV

Area of the square loop (A) = $(4\times {10}^{(}-2){)}^2=1.6\times {10}^{-3}{m}^2$. Magnetic flux $\varphi =ABcos\theta ,whereBcos\theta$ is the component of the magnetic field perpendicular to the plane of the loop, Therefore, change in flux is
$\Delta \varphi =ABcos\theta =1.6\times {10}^{-3}\times 0.5\times cos{60}^o=0.4\times {10}^{-3}Wb\therefore Inducedemfe=\frac{\Delta \varphi }{\Delta t}=\frac{0.4\times {10}^{-3}}{0.2}=2\times {10}^{-3}V=2mV$.