Question

In a region of space where the magnetic field of the earth has a magnitude of 80 μT and is directed 30° below the horizontal, a 50-cm length of wire oriented horizontally along an east-west direction is moved horizontally to the south with a speed of 20 m/s. What is the magnitude of the induced potential difference between the ends of this wire?

Answer 1

V_ind = 4 × 10^(-4) V

Step-by-step explanation:

We are given;

Magnetic field; B = 80 μT = 8 × 10^(-5) T

Angle;θ = 30°

Lenght;L = 50 cm = 0.5 m

Speed; v = 20 m/s

Now, formula for the induced potential difference is known as;

V_ind = NBLVsin θ

Where;

V_Ind = Induced potential difference/voltage

N = Number of turns

B = Magnetic field

V = velocity

L = length

Number of turns in this case is 1 since it's just a wire between both ends.

Thus, plugging in the relevant values, we have;

V_ind = 1 × 8 × 10^(-5) × 20 × 0.5 × sin 30

V_ind = 4 × 10^(-4) V