A solenoidal coil with 25 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 23.0cm long and has a dia

Question

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 23.0cm long and has a diameter of 2.00cm . At a certain time, the current in the inner solenoid is 0.150A and is increasing at a rate of 1600A/sPart AFor this time, calculate the average magnetic flux through each turn of the inner solenoid.Part BFor this time, calculate the mutual inductance of the two solenoids;Part CFor this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

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Thiên Thanh 3 months 2021-07-19T16:03:10+00:00 2 Answers 2 views 0

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    0
    2021-07-19T16:04:21+00:00

    Answer:

    see the answers below please

    Explanation:

    We have that the current in the inner solenoid is:

    i(t)=0.150+1600t

    A) the magnetic flux is given by

    \Phi_B=BS=\pi r^2 B

    B is obtained by computing for a solenoid:

    B=\frac{\mu_0n_1i(t)}{L}

    hence, we have

    \Phi_B=\frac{\pi r^2 \mu_0 n_1}{L}(0.150+1600t)

    B)

    the mutual inductance is obtained by using:

    M_{12}=M_{21}=M=\mu_0n_1n_2\pi r_{1}^2i(t)=\pi \mu_0 r^2 (25)(320)(0.150A+1600t)

    C)

    emf=-\frac{d\Phi_B}{dt}=-\frac{\pi r_2^2\mu_0n_2}{L}(1600)

    hope this helps!!

    0
    2021-07-19T16:04:28+00:00

    Answer: Part A, magnetic flux = 9.02 × 10^-8 Wb

    Part B, mutual inductance, M = 1.50 × 10^-5H

    Part C, emf induced in the outer solenoid = -0.024V

    Explanation: Please see the attachments below

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