Question

My Notes A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.20 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 1.30 ?

Answer 1

0.00158 A

Step-by-step explanation:

A = Area of loop = 7.8 cm²

`B_f` = Final magnetic field = 3.2 T

`B_i` = Initial magnetic field = 0.5 T

R = Resitance =
`1.3\ \Omega`

The emf is given by

`E=((B_f-B_i)A)/(t)\\\Rightarrow E=((3.2-0.5)* 7.8* 10^(-4))/(1.02)\\\Rightarrow E=0.00206\ V`

The current is given by

`I=(E)/(R)\\\Rightarrow I=(0.00206)/(1.3)\\\Rightarrow I=0.00158\ A`

The resulting induced current in the loop is 0.00158 A