Question

A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diameter 1.25 cm, and resistance of 3.25 µΩ, lies stationary in the same plane as the straight wire. At some initial time, the wire is at a distance d = 20.0 cm from the center of the coil. 4.70 s later, the wire is at a distance 2d from the center of the coil. What is the magnitude and direction of the average induced current in the coil? Note that while the magnetic field varies over the diameter of the coil, it is very small and we will disregard this variation.

Answer 1

Answer:

I = 69.3  μA

Explanation:

Current through the straight wire, I = 3.45 A

Number of turns, N = 5 turns

Diameter of the coil, D = 1.25 cm

Resistance of the coil, $R = 3.25 \mu ohms$

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

$B_{1} = \frac{\mu_{0}I }{2\pi d}$

$B_{1} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T$

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

$B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d }$

$B_{2} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T$

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, $I = \frac{E}{R}$

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ =  -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

$E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V$

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

$I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }$

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3  μA