Question

A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10-s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. (a) Determine the emf induced in the loop. (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop

Answer 1

Answer:

ε= 7.86 mV,  Current: Anti-clockwise

Explanation:

radius= 50 mm

dt= 0.10 s

Initial magnitude of magnetic field= B1 = 200 mT

Final magnitude of magnetic field = B2 = 300 mT

Ф= B. A= BA cosα

Ф1= B1 * A * cosα

Ф1= $(200*10^{-3})* \pi * (50*10^{-3} )^2*(1)$

Ф1= 0.00157 Wb

Ф2= B2 * A * cosα

Ф1= $(300*10^{-3})* \pi * (50*10^{-3} )^2*(1)$

Ф2=0.00236 Wb

dФ= Ф2 - Ф1

dФ= 0.00236 - 0.00157

dФ= 0.000786 Wb

ε= $\frac{d}{dt}$ Ф

ε=0.001786/ 0.10

ε=0.00786 v = 7.86 mV

b)

According to lenz's law the induced emf always oppose the cause producing it.

Applied field is out of the paper so the current will flow in anti-clockwise direction to produce north pole pointing toward the paper.