Question

A square wire loop of side length a = 2.0 cm and resistance R = 10.0Ω is inside a 1.00m long solenoid with 1000 windings such that the plane of the loop is perpendicular to the solenoid’s magnetic field. The solenoid initially carries a current I = 6.0 A, which is turned down to zero evenly over a period of 12.0 s. If the solenoid’s initial magnetic field is out of the page in the figure, what is the magnitude and direction of the current in the square loop as the current in the solenoid is turned off?

Answer 1

Given

Side length of the square loop                l = 2.0 x 10^-2 m

Resistance of the square loop                     R = 10.0 Ω

Length of the solenoid                                 L = 1.00 m

Number of windings                                                 N = 1000

Magnitude of current at the beginning       I_o = 6 A

Magnitude of current at the end                 I_f = 0 A

Period of change                                           Δt = 12.0 s

Solution

Area of the square loop

A = l ²

A = (2.0x10^-2)²

A = 4.0 x 10^-4 m²

Since, flux is the number of lines of magnetic induction passing through a enclosed surface area

The magnetic flux of the solenoid enclosed in this surface

Φ = µ₀N²AI/L

The change in current will create a change in flux which in turn will create an emf in the square loop

emf ɛ = -ΔΦ/Δt

emf ɛ = -µ₀N²A ΔI/L Δt

emf ɛ = -4π x 10^-7 x 1000² x 4.0 x 10^-4 (I_f - I_o)/ 1 x 12

emf ɛ = - 4.1867 x 10^-5 (0-6

emf ɛ = 25.13 V

The current flowing in the square loop

I_wire = ɛ/R

I_wire = 25.13/10

I_wire = 2.513 A

This current flows in anti-clock wise since the induced magnetic field is acting out of page