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A 431-turn solenoid with a length of 28 cm and radius of 1.4 cm carries a curren

ID: 2018201 • Letter: A

Question

A 431-turn solenoid with a length of 28 cm
and radius of 1.4 cm carries a current of 2.2 A.
A second coil of 4 turns is wrapped tightly
about this solenoid so that it can be consid-
ered to have the same radius as the solenoid.
Find the change in the magnetic flux
through the coil when the current in the
solenoid increases to 6.8 A in a period of
1.17 s. The permeability of a vacuum is
4 pi × 10-7 T · m/A
Answer in units of T · m2.

Find the magnitude of the average induced
emf in the coil.
Answer in units of µV.

Explanation / Answer

            The magnetic field   due to solenoid is     B = 0nI    ...........(1)    The induced emf in the second coil      = N   d /dt                       =   N d(BA) /dt     ...........(2) Here N = number   of tirns of the secon coil      plug (1)   in (2)                   =   0nNA(dI /dt )   .................(3)         Given   0 =   4 *10-7 H/m               N =   4 turns ,      n = number of turns of the solenoid per unit length =   431 /0.28 = 1539.28 turns /m     A   = srea of the secon coil   = r2 = (0.014 m)2 = 0.00061544 m2 dI / dt    =   ( 6.8 -2.2 ) A / 1.17s   =   3.93 A / s      plug all values in (3) we get            induced emf     =   18.7*10-6 V   = 18.7 V
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