Question

5 single loop of copper wire, lying flat in a plane, has an area of 8.60cm2 and a resistance of 3.00. A uniform magnetic field points perpendicular to the plane of the loop. The field initially has a magnitude of .5 T, and the magnitude increases linearly to 3. 0 in a time of 1.10 s, what is the induced current in the loop of wire over this time? .652A

Answer 1

Answer:

0.652 mA

Explanation:

According to Faraday's Law :

$Emf = - \frac{d \phi}{dt}$

$= \frac{- \delta \phi }{\delta \ t}$

$|E| = \frac{A ( \delta B)}{\delta t}$

where ;

A = $8.60*10^{-4}$

$\delta B = 3.00 T - 0.5 T = 2.5 T$

$|E| = \frac{8.60*10^{-4}*2.5}{1.10}$

$|E| = 1.96*10^{-3}$

Induced current  I = $\frac{|E|}{R}$

= $\frac{1.96*10^{-3}}{3.0}$

= $6.52*10^{-4} A$

= 0.652 mA

Thus, the induced current in the loop of wire over this time = 0.652 A