Question

A closed loop conductor with radius 2.0 m is located in a changing magnetic field. If the maximum emf induced in the loop is 2.0 V, what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies?

Answer 1

The maximum rate at which the magnetic field strength is changing is determined using Faraday's law and is calculated to be 0.159 T/s.

Step-by-step explanation:

To calculate the maximum rate at which the magnetic field strength is changing, we can use Faraday's law of electromagnetic induction, which in its differential form states that the electromotive force (emf) induced in a loop is equal to the negative change in magnetic flux through the loop per unit time: ε = - dΦB/dt, where ε is the induced emf, and ΦB is the magnetic flux. The magnetic flux (ΦB) through a circular loop of radius r with a magnetic field B perpendicular to the loop's plane is given by ΦB = B × (pi × r2).

Since the problem provides the maximum emf (ε = 2.0 V) and the radius of the loop (r = 2.0 m), we can set up the equation for the rate of change of the magnetic field strength as: ε = - d(B × (pi × r2))/dt = - pi × r2 × dB/dt

By rearranging the equation to solve for dB/dt and substituting the values we have, we get:
dB/dt = -ε / (pi × r2)
dB/dt = -(2.0 V) / (pi × (2.0 m)2)
dB/dt = -0.15915 T/s

The negative sign indicates that the magnetic field is decreasing, but since the question asks for the magnitude of the rate of change, we can provide the answer as 0.159 T/s as the maximum rate of change of the magnetic field strength.

Answer 2

0.159 Tesla per second

Step-by-step explanation:

r = 2 m, e = 2 V,

Let the rate of magnetic field changing is dB / dt.

By use of Farady's law of electromagnetic induction

e = rate of change of magnetic flux

e = dФ /dt

e = A x dB / dt

2 = 3.14 x 2 x 2 x dB / dt

dB / dt = 0.159 Tesla per second