session.masteringphysics.com | Constants Part A A very long, straight solenoid w
ID: 2040845 • Letter: S
Question
session.masteringphysics.com | Constants Part A A very long, straight solenoid with a cross-sectional area of 1.88 cm is wound with 88.7 turns of wire per centimeter. Starting at t-0, the current in the solenoid is increasing according to i(t) 0.176 A/s2 )t2. Asecondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A? Express your answer with the appropriate units. IEI- 1 Value Units Submit Request Answer Provide Feedback Next>Explanation / Answer
We know that
Emf=N*A*(dB/dt)
where, N = number of turns
N = 88.7 turns/cm = 8870 turns/metre
A = area of solenoid
The field inside the long solenoid is given by the relation
B = ??ni
where
i =i(t) = (0.176) t2
implies that
B = (4?x10-7) x 8870 x 0.176t² = 1.96 x 10-3 t²
The rate of change of magnetic field is thus
dB/dt = 3.92 x 10-3 t
A = 1.88 cm2 = 1.88 x 10-4 m2
the emf is thus
|Emf| = rate of change of flux linkage
|Emf| = d(NAB)/dt = NA dB/dt
= 5 x (1.88 x 10-4) x (3.92 x 10-3 t)
= 3.68 x 10-6
let T be the time at which the current I = 3.2 A,then
3.2 = 0.176 T²
which gives the value of T = 4.264 s
thus emf is
|emf| = 3.68 x 10-6 T
= 3.68 x 10-6 x (4.264)
= 1.57 x 10-5 V
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