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When the current in a long, straight air-filled solenoid is changing at the rate of 3000 A/S, the voltage across the solenoid is 0.600 V. The solenoid has 1200 turns and uniform cross-sectional area 35.0 mm? Assume that the magnetic field is uniform inside the solenoid and zero outside, so the inductance formula L MAN/ 2/1 for a solenoid with N turns, uniform cross-sectional area A, and length 1, applies Part A What is the magnitude B of the magnetic field in the interior of the solenoid when the current in the solenoid is 3.00 A? Express your answer with the approppriate units. μΑ ? B= Value Units Submit Request Answer

Respuesta :

The magnitude of B in the interior of the solenoid is 5 * 10⁻⁷ A/m.

Given that,

Rate of change of current = 3000 A/s

EMF induced ε = 0.6 V

Number of turns N = 1200 turns

Cross-sectional area A = 35 mm² = 35 * 10⁻⁶ m²

Current i = 3 A

We know the relation, ε = L* dI/dt

Making L as subject, L = ε / ( dI/dt )

Substituting the values, we have

L = 0.6 / 3000 = 0.0002 H

We know the relation between N, i, B and L as

N* B = L *i

Making B as subject, we have

B = L *i/N = (0.0002 * 3)/1200 = 0.0000005 A/m = 5 * 10⁻⁷ A/m

Thus, magnetic field in the interior of the solenoid is 5 * 10⁻⁷ A/m.

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