A solenoid of radius 2 cm and length 15 cm has 120 turns.The rate of change of induced current through it to produce a self-induced emf of 4 mV is
A)2.3 A/s.
B)8.5 A/s.
C)12.9 A/s.
D)26.3 A/s.

Question

Quizplus · Accepted Answer

Given, 
radius of the solenoid, r = 2 cm = 0.02 m 
length of the solenoid, l = 15 cm = 0.15 m 
number of turns, N = 120 
self-induced emf, ε = 4 mV = 4 × 10⁻³ V

Using the formula for self-induced emf in a solenoid,

ε = -N(dΦ/dt)

where, dΦ/dt is the rate of change of magnetic flux.

The magnetic flux through a solenoid is given by,

Φ = μ₀NLi

where, 
μ₀ = 4π × 10⁻⁷ Tm/A is the permeability of free space 
i is the current in the solenoid

Differentiating with respect to time, we get

dΦ/dt = μ₀Nl(d/dt)i

Substituting this in the expression for self-induced emf, we get

ε = -μ₀N²Al(d/dt)i

where, A = πr² is the cross-sectional area of the solenoid.

Solving for (d/dt)i, we get

(d/dt)i = -ε/(μ₀N²Al)

Substituting the given values, we get

(d/dt)i = -(4 × 10⁻³)/(4π × 10⁻⁷ × 120² × π × (0.02)² × 0.15) = -26328.6 A/s ≈ -26.3 A/s

The negative sign indicates that the induced current will flow in the opposite direction to the change in magnetic flux that produced it. Therefore, the rate of change of induced current required to produce a self-induced emf of 4 mV is 26.3 A/s, which is closest to option D) 26.3 A/s.

A Solenoid of Radius 2 Cm and Length 15 Cm