A solenoid of radius 2 cm and length 15 cm has 120 turns.If the rate of change of induced current through it is 12.9 A/s,what is the self-induced emf?
A)2.71 mV
B)3.41 mV
C)1.96 mV
D)4.00 mV

Question

Quizplus · Accepted Answer

The self-induced emf can be calculated using the formula: 
emf = -N(dΦ/dt), where N is the number of turns, and dΦ/dt is the rate of change of magnetic flux. 
The magnetic flux through a solenoid is given by: Φ = μ0*N*I*A, where μ0 is the permeability of free space, I is the current, and A is the cross-sectional area of the solenoid. 
We can find A using the radius and length given: A = π*r^2 = π*(0.02 m)^2 = 1.26x10^-3 m^2. 
Substituting these values, we get: 
emf = -N*d(μ0*N*I*A)/dt 
= -N*μ0*N*A*dI/dt 
= -(120)*(4πx10^-7)*(120)*(1.26x10^-3)*(12.9) 
= 1.96x10^-3 V = 1.96 mV. 
Therefore, the answer is C.

A Solenoid of Radius 2 Cm and Length 15 Cm