A solenoidal coil with 26 turns of wire is wound tightly around another coil wit
ID: 1512177 • Letter: A
Question
A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.10 cm . At a certain time, the current in the inner solenoid is 0.110 A and is increasing at a rate of 1500 A/s .
a)For this time, calculate the average magnetic flux through each turn of the inner solenoid.
b)For this time, calculate the mutual inductance of the two solenoids;
M=?
C)For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.
Explanation / Answer
magnetic fild due to inner solenoid is
B = uo ( N2/l) i
= 4 pi * 10^-7 ( 350/0.25) ( 0.110)
=1.93 * 10^-4 T
magnetic flux is
flux = BA = 1.93 * 10^-4 T ( pi (0.021/2)^2 = 6.68 * 10^-8 Wb
---------------------------------------------------------------------------
M = N flux/ i1
= 26 ( 6.68 * 10^-8 Wb)/0.110=1.57 * 10^-5 H
(c)
e = - M di2/dt
= - ( 1.57 * 10^-5 H) (1500 A/s)
=-0.02355 V
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