Question

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume that the loop is perpendicular to the magnetic field. What is the average induced emf?

Answer 1

9.88 milivolt

Step-by-step explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf = - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

`=-\pi(d/2)^2(B_2-B_1)/(\Delta t) \\=-\pi(0.052/2)^2(0-1.35)/(0.29) \\=98.8*10^4\\=9.88 mV`