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

Answer:

0.00158 A

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

$E=\frac{(B_f-B_i)A}{t}\\\Rightarrow E=\frac{(3.2-0.5)\times 7.8\times 10^{-4}}{1.02}\\\Rightarrow E=0.00206\ V$

The current is given by

$I=\frac{E}{R}\\\Rightarrow I=\frac{0.00206}{1.3}\\\Rightarrow I=0.00158\ A$

The resulting induced current in the loop is 0.00158 A