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 ((dB)/(dt))`

Here N =1

`\epsilon_(emf) = (1)(7.00 x 10^-4 m2)[((3.3 T - 0.5 T))/((0.99s))]`

`\epsilon_(emf)= 0.00196 V`

b) The magnitude of the induced current is

`I = (V)/(R)`

Here Resistance is

`R = 1.80\Omega`

`I = ((0.00196V))/((1.80\Omega))`

`I = 0.00108 A`

Therefore the induced current is 0.00108A