Question

A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increases uniformly inmagnitude from 0.500 T to 3.50 T in1.00 s. What is the resulting inducedcurrent if the loop has a resistance of 1.30 Ω?

Answer 1

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

$\Phi = BA Cos\theta$

Where

B= Magnetic Field

A = Area

$\theta =$ Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

$\theta =$ 0 then our expression can be written as

$\Phi = BA$

From the same value of the electromotive force we have to

$\epsilon = -\frac{d\Phi}{dt}$

Replacing we have

$\epsilon = -A\frac{B}{dt}$

Replacing with our values we have that

$\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}$

$\epsilon = -0.0237V$

Therefore the magnitude of the induced emf in the loop is 0.0237V

On the other hand we have that the current by Ohm's Law can be defined as

$I = \frac{\epsilon}{R}$

For the given value of the resistance and the previously found potential we have to

$I = \frac{0.237}{1.3}$

$I= 0.0182A$