Chapter 22, Problem 19 A rectangular loop of wire with sides 0.220 and 0.407 m l

Question

Chapter 22, Problem 19 A rectangular loop of wire with sides 0.220 and 0.407 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field ha a magnitude of 0.673 T and is directed parallel to the normal of the loop's surface. In a time of 0.252 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop. xxX (al Number the tolerance is +/-3% Click if you would like to Show WorlA Units V/m uestion: Wb rad/s

Explanation / Answer

Problem 19) We know that emf induced is change in magnetic flux

e=-d/dt

where =BA since area vector and magnetic field vectors are parralel here

A=0.220*0.407=0.08954 m2

B=0.673 T

Now area is reduced to half Anew=A/2

change in time=0.252 sec

e=-d/dt

e=- B * d A/dt

e= -(0.673) * (A/2-A)/dt

e=(0.673 * A)/2 *0.252

e=(0.673 * 0.08954)/(2 *0.252)

e=0.11956 V

Problem 67) P=1.86 * 106 W

Total resistance=R=4.9 * 10-2 *5.4= 0.2646 ohms

I=P/V=1.86 * 106/1600

I=1125 A

Plost=I2R=1125 *1125 *0.2646=334884.375 W

ii)   The current will be reduced by a factor of 1/100, so the power lost in the wires will be reduced by a factor of 10^-4,

Plost new= Plost * 10-4=33.4884 W

GOOD LUCK!!!

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