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

(a) 16.5528 V

(b) 827.64 W

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

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when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.