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 cmand 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 VRelated Questions
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