Question

The magnetic field perpendicular to a single 17.2-cm-diameter circular loop of copper wire decreases uniformly from 0.590 T to zero.

If the wire is 2.75 mm in diameter, how much charge moves past a point in the coil during this operation? The resistivity of copper is 1.68×10−8Ω⋅m.

Answer 1

Given

Diameter of the circular loop D_L = 17.2 cm = 0.172 m

Diameter of the wire D_w = 2.75 mm = 2.75 x 10^-3 m

Resistivity of copper ρ= 1.68 x 10^-8 ohm.m

Change in magnetic field ΔB = 0 - 0.590 = -0.590 T

(let’s say the time taken for the filed to go through this change is Δt )

The angle between the surface vector and the field θ = 0^o

(the surface vector is perpendicular to surface but parallel to magnetic field)

Solution

Area of the loop

A_L = πr_L² = π(D_L/2)² = 3.14 x (0.172/2)² = 0.02322344 m²

Emf induced in the loop

e = -Acosθ (ΔB/Δt)

e = - 0.02322344 x 1 x (-0.590/Δt)

e = 0.0137018296/Δt

e = IR

0.0137018296/Δt = IR

0.0137018296/Δt = (Δq/Δt)R

0.0137018296= (Δq)R

0.0137018296= (Δq)ρL/A_w

The length of the wire L = 2πr_L

Area od cross section of the wire A_w = πr_w²

0.0137018296= (Δq)ρ2πr_L/πr_w²

0.0137018296= (Δq)ρ2r_L/r_w²

0.0137018296= (Δq)ρD_L/r_w²

0.0137018296= (Δq) x 1.68 x 10^-8 x 0.172 / (2.75 x 10^-3/2)²

Δq = 8.96 C