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A 420-turn circular coil with an area of 0.0650 m2 is mounted on a rotating frame, which turns at a rate of 22.3 rad/s in the presence of a 0.0550-T uniform magnetic field that is perpendicular to the axis of rotation. What is the instantaneous emf in the coil at the moment that the normal to its plane is perpendicular to the field

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1 vote

Answer:

33.48 V

Step-by-step explanation:

Parameters given:

Number of turns, N = 420

Magnetic field strength, B = 0.055 T

Area, A = 0.065 m²

Angular velocity, ω = 22.3 rad/s

EMF induced in a coil is given as:

EMF = -dΦ/dt

where Φ = magnetic flux

Magnetic flux, Φ, is given as:

Φ = B * N * A * cosωt

EMF = -d( B * N * A * cosωt) / dt

EMF = B * N * A * ω * sinωt

where ωt = 90°

Therefore:

EMF = 0.055 * 420 * 0.065 * 22.3 * sin90°

EMF = 33.48 V

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