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Problem 20.75 A thin rectangular coil 2 cm by 9 cm has 44 turns of copper wire. It is made to rotate with angular frequency 110 rad/s in a magnetic field of 1.9 T. (a) What is the maximum emf produced in the coil? V (b) What is the maximum power delivered to a 50 ohm resistor? W

User RSK
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1 Answer

1 vote

Answer:

(a) 16.5528 V

(b) 827.64 W

Step-by-step explanation:

(a)

The formula for maximum emf produced in a coil is given as

E₀ = BANω.................. Equation 1

Where E₀ = maximum emf produced in the coil, B = magnetic Field, A = Area of the rectangular coil, N = number of turns, ω = angular velocity.

Given: B = 1.9 T, N = 44 turns, ω = 110 rad/s, A = 2×9 = 18 cm² = 18/10000 = 0.0018 m²

Substitute into equation 1

E₀ = 1.9(44)(110)(0.0018)

E₀ = 16.5528 V

(b)

Maximum power

P₀ = E₀²/R...................... Equation 2

Where R = Resistance.

Given: R = 50 ohms, E₀ = 16.5528 V

Substitute into equation 2

P₀ = 50(16.5528)

P₀ = 827.64 W

User KylePDavis
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