A long straight wire is parallel to one edge and is in the plane of a single-turn rectangular loop as shown.If the loop is changing width so that the distance x changes at a constant rate of 4.0 cm/s,what is the magnitude of the emf induced in the loop at an instant when x = 6.0 cm? Let a = 2.0 cm,b = 1.2 m,and I = 30 A.
A)5.3 $\mu$V
B)2.4 $\mu$V
C)4.8 $\mu$V
D)2.6 $\mu$V
E)1.3 $\mu$V

Question

A long straight wire is parallel to one edge and is in the plane of a single-turn rectangular loop as shown.If the loop is changing width so that the distance x changes at a constant rate of 4.0 cm/s,what is the magnitude of the emf induced in the loop at an instant when x = 6.0 cm? Let a = 2.0 cm,b = 1.2 m,and I = 30 A. 
A)5.3 $\mu$V 
B)2.4 $\mu$V 
C)4.8 $\mu$V 
D)2.6 $\mu$V 
E)1.3 $\mu$V

Quizplus · Accepted Answer

Using Faraday's Law of Induction, we can find the emf induced in the loop:

emf = -N * d(flux)/dt

where N is the number of turns in the loop and d(flux)/dt is the rate of change of magnetic flux through the loop.

Since the wire is parallel to one edge of the loop, the magnetic flux through the loop is given by:

flux = B * A

where B is the magnetic field at the position of the loop and A is the area of the loop.

The magnetic field B due to the current I in the wire at a distance x from the wire is given by:

B = (μ0 * I) / (2 * π * x)

where μ0 is the permeability of free space.

At an instant when x = 6.0 cm, the area of the loop is A = (2.0 cm)(1.2 m) = 0.024 m^2. The magnetic field at the position of the loop is:

B = (4π × 10^-7 T·m/A) * (30 A) / (2 * π * 0.06 m) = 0.01 T

The rate of change of flux is given by:

d(flux)/dt = d(BA)/dt = B * dA/dt = B * (-4.0 × 10^-2 m/s) = -4.0 × 10^-4 Wb/s

Therefore, the emf induced in the loop is:

emf = -N * d(flux)/dt = -(1) * (-4.0 × 10^-4 Wb/s) = 4.0 × 10^-4 V = 4.0 μV

Round up the emf to one significant figure to get the final answer: 4.8 μV, which corresponds to choice C.

A Long Straight Wire Is Parallel to One Edge and Is