A planar loop consisting of seven turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field that increases uniformly in magnitude from 14 * 10-3 T to 38 * 10-3 T in a time of 8.0 * 10-3 s. What is the resulting induced current in the coil if the resistance of the coil is 5.0 Ω?

The induced current is 0.084 A

Explanation:

the area given by the exercise is

A = 200 cm^2 = 200x10^-4 m^2

R = 5 Ω

N = 7 turns

The formula of the emf induced according to Faraday's law is equal to:

ε = (-N * dφ) / dt = (N * (b2-b1) * A) / dt

Replacing values:

ε = (7 * (38 - 14) * (200x10^-4)) / 8x10^-3 = 0.42 V

the induced current is equal to:

I = ε / R = 0.42/5 = 0.084 A

Given Information:

Area of loop = A = 200 cm² = 0.0200 m²

Change in time = Δt = 8x10⁻³ seconds

Change in magnetic field = ΔB = (38x10⁻³ - 14x10⁻³) T

Number of turns = N = 7

Resistance of coil = R = 5 Ω

Required Information:

Induced current = I = ?

Answer:

Induced current = 0.084 A

Explanation:

From the Faraday's law the induced EMF ξ in the coil is given by

ξ = - NΔΦ/Δt

Where ΔΦ is the change in flux

ΔΦ = ΔBA

Where A is the area of the planer loop

ΔΦ = (38x10⁻³ - 14x10⁻³) * 0.0200

ΔΦ = 0.00048

So the induced emf becomes

ξ = - NΔΦ/Δt

ξ = (-7*0.00048) / 8x10⁻³

ξ = - 0.42 V

The negative sign indicates that the induced emf opposes the change that produced it in the first place.

Finally, we can now find the induced current using Ohm's law

I = ξ/R

I = 0.42/5

I = 0.084 A

Therefore, the resulting induced current in the coil is 0.084 A