Question

A flat loop of wire consisting of a single turn of cross-sectional area 7.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.80?

Answer 1

To solve this problem we must first find the potential on the body which is given as a product between the number of turns, the area and the variation of the magnetic field as a function of time. Once the potential is found, we will apply Ohm's Law with which we can find the induced current on the body. Our values are,

$A = 7.00*10^{-4} m^2$

$\Delta B = 3.30T-0.5T$

$t = 0.99s$

$R = 1.8\Omega$

a) The magnitude of average induced emf is given by

$\epsilon_{emf} = NA (\frac{dB}{dt})$

Here N =1

$\epsilon_{emf} = (1)(7.00 x 10^-4 m2)[\frac{(3.3 T - 0.5 T)}{(0.99s)}]$

$\epsilon_{emf}= 0.00196 V$

b) The magnitude of the induced current is

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

Here Resistance is

$R = 1.80\Omega$

$I = \frac{(0.00196V)}{(1.80\Omega)}$

$I = 0.00108 A$

Therefore the induced current is 0.00108A