An 8.50 cm long solenoid has 110 loops with a radius of 2.10 cm. A single metal

Question

An 8.50 cm long solenoid has 110 loops with a radius of 2.10 cm. A single metal loop of resistance 3.90 Ohms, whose radius is twice that of the solenoid, is centered around the solenoid as shown to the right. The current through the wire of the solenoid increases uniformly from 0.88 A to 1.00 A over 2.50 s and flows clockwise around the solenoid, as seen when looking at the left end of the solenoid from the left. What average cmf is induced around the loop? What is the direction of the current around the loop as seen from the left side? Fully explain your reasoning.

Explanation / Answer

n=no of turns per unit length=110/0.085=1294.1

B due to solenoid inside it =0ni

flux through the loop = integral B.ds .

note B due to a solenoid (very long) is zero outside it. so the flux through the outer loop will have that area only which have non ZERO value of B. also the field inside solenoid is uniform and hence it will come out of the integral sign.

flux through the loop = integral B.ds =0ni**(0.021)2

=22.50i

average EMF= -/T=-22.50i/T

= -22.50*0.048

= -1.08V

PART 2

since the current is increasing hence the flux is increasing so EMf induced will be such that the current produced due to this EMF cancels the increased current.IF INITIALLY FROM THE LEFT SIDE IF THE CURRENT IN THE SOLENOID WAS IN CLOCKWISE DIRECTION THEN INDUCED CURRENT WILL BE COUNTER-CLOCKWISE DIRECTION AND VICE-VERSA.

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