Question

A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. (a) What is the magnitude of the change in the external magnetic flux enclosed by the coil? (b) What is the magnitude of the average voltage induced in the coil as the external flux is changing? (c) If the coil has a resistance of 4.0 ohms, what is the magnitude of the average current in the coil?

Answer 1

(a) 0.0068 Wb

Since the plane of the coil is perpendicular to the magnetic field, the magnetic flux through the coil is given by

$\Phi = NBA$

where

N = 200 is the number of loops in the coil

B is the magnetic field intensity

$A=8.5 cm^2 = 8.5\cdot 10^{-4} m^2$ is the area of the coil

At the beginning, we have

$B_i = 0.060 T$

so the initial magnetic flux is

$B_i = (200)(0.060 T)(8.5\cdot 10^{-4} m^2)=0.0102 Wb$

at the end, we have

$B_f=0.020 T$

so the final magnetic flux is

$B_f = (200)(0.020 T)(8.5\cdot 10^{-4} m^2)=0.0034 Wb$

So the magnitude of the change in the external magnetic flux through the coil is

$\Delta \Phi = |\Phi_f - \Phi_i|=|0.0034 Wb-0.0102 Wb|=0.0068 Wb$

(b) 0.567 V

The magnitude of the average voltage (emf) induced in the coil is given by Faraday-Newmann law

$\epsilon= \frac{\Delta \Phi}{\Delta t}$

where

$\Delta \Phi = 0.0068 Wb$ is the variation of magnetic flux

$\Delta t = 12 ms = 0.012 s$ is the time interval

Substituting into the formula, we find

$\epsilon=\frac{0.0068 Wb}{0.012 s}=0.567 V$

(c) 0.142 A

The average current in the coil can be found by using Ohm's law:

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

where

I is the current

V is the voltage

R is the resistance

Here we have:

V = 0.567 V (induced voltage)

$R=4.0 \Omega$ (resistance of the coil)

Solving for I, we find

$I=\frac{0.567 V}{4.0 \Omega}=0.142 A$