What is the magnitude of the rate of change of the magnetic field inside the coil? You will need to calculate this algebraically before you can get a number. Write an expression for the magnetic field due to the wire at the location of the coil. Use the approximate formula, since the wire is very long. Remember the chain rule, and remember that ⅆ ⅆ .

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ogbe2k3 ogbe2k3 · Answer 1 · 2020-05-04T15:47:33+02:00

Answer:

The answer to this questions are, (a) 0.685 * 10^-4 T/sec (b) 0.946 * 10^-6V

Explanation:

Solution

Recall

radius r = 0.02m

N = 11 turns

Instant I = 3 amperes

Velocity v =3.3m/s

x = 0.17 m

(a) What is the magnitude of the rate of change of the magnetic field inside the coil

db/dt =μ₀T/2π x²

Thus,

= 4π * 10^-7* 3* 3.3/2π * 0.17²

=685.12* 10^-7

which is now,

0.685 * 10^-4 T/sec

(b) What is the magnitude of the voltmeter reading.

μ = N (db/dt)πr²

Note: this includes all 11 turns of the coil

Thus,

= 11 * 0.6585* 10^-4* 3.14 * (0.02)²

= 946 * 10^-7

which is = 0.946 * 10^-6V

Note: Kindly find an attached copy or document of the complete question of this exercise