Question

A conducting circular loop of radius $\frac{10}{\sqrt{\pi}}$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.05 s is :

• A. emf = 1 mV
• B. emf = 5 mV
• C. emf = 10 mV
• D. emf = 100 mV

Answer 1

Answer: (C)

emf = 10 mV

Answer 2

The induced emf in a loop is given by Faraday's Law: $\mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t}$. For a single loop ($N=1$), the magnetic flux is $\Phi_B = B \cdot A$. The radius of the loop is $r = \frac{10}{\sqrt{\pi}}$ cm $= \frac{10}{\sqrt{\pi}} \times 10^{-2}$ m. The area of the loop is $A = \pi r^2 = \pi \left(\frac{10}{\sqrt{\pi}} \times 10^{-2}\right)^2 = 10^{-2}$ m$^2$. The magnetic field changes from $B_i = 0.5$ T to $B_f = 0$ T in $\Delta t = 0.5$ s. The change in magnetic flux is $\Delta \Phi_B = (B_f - B_i) A = (0 - 0.5) \times 10^{-2}$ Wb $= -0.5 \times 10^{-2}$ Wb. Since the magnetic field is decreased at a steady rate, the induced emf is constant. The magnitude of the induced emf is $\mathcal{E} = \left| -1 \times \frac{-0.5 \times 10^{-2} \text{ Wb}}{0.5 \text{ s}} \right| = 10^{-2}$ V $= 10$ mV.