A loop of wire of radius a = 50. mm has an electrical resistance R = 0.031 ? . T

Question

A loop of wire of radius a = 50. mm has an electrical resistance R = 0.031 ? . The loop is initially inside a uniform magnetic field of magnitude B0 = 1.3 T parallel to the loop's axis. The magnetic field is then reduced slowly at a constant rate, which induces a current I = 0.20 A in the loop. How long does it take for the magnitude of the uniform magnetic field to drop from 1.3 T to zero?

Part B: What is the direction of the induced current in the wire shown in the picture?

A. Clockwise

B. Counterclockwise

* I AM MOST CONCERNED FOR THE ANSWER FOR PART B. IS IT CLOCKWISE OR COUNTERCLOCKWISE?*

Explanation / Answer

part a:

let time taken for the uniform magnetic field to drop from 1.3 T to 0 is t seconds.

then magnitude of rate of change of magnetic flux density=(1.3-0)/t=1.3/t T/s
induced emf=number of turns*rate of change of flux linkage

==>induced current*resistance=1*rate of change of (magnetic flux density*area of the coil)

==>0.2*0.031=(1.3/t)*pi*0.05^2

==>t=1.3*pi*0.05^2/(0.2*0.031)=1.6468 seconds

part b:

according to lenz's law:

the direction of induced current is such that it opposes the original field that created the induced current

initially the field was into the plane of the paper.

let it be -ve z direction

hence the magnitude of magnetic field is decreasing in -ve z direction

so the induced field due to the induced current should be in -ve z direction as per the lenz's law.

(consider it this way..if field is decreasing in -ve z direction, it is increasing in +ve z direction

hence the field due to induced current should be such that it opposes that increment

hence field should be along -ve z direction)

now, using right hand thumb rule,

we can see that a clockwise current will produce a magnetic field in -ve z direction

hence option A is correct.

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