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worty [1.4K]
3 years ago
11

An elastic conducting material is stretched into a circular loop of 14.7 cm radius. It is placed with its plane perpendicular to

a uniform 0.963 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 74.5 cm/s. What emf is induced in volts in the loop at that instant?
Physics
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer: 0.66 V

Explanation:

Given

Magnetic field, B = 0.963 T

Instantaneous rare = 74.5 cm/s = 0.745 m/s

radius, r = 14.7 cm = 0.147 m

We will use the formula

emf = dΦ/dt

emf = d(BA)/dt

emf = d(Bπr²)/dt

if B is constant, then we can say

emf = Bπ d(r²)/dt on differentiating, we have,

emf = Bπ (2r dr/dt)

emf = 2πrB dr/dt substituting each values, we have

emf = 2 * 3.142 * 0.147 * 0.963 * 0.745

emf = 0.66 V

Therefore, the induced emf in the loop at that instant is 0.66 V

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Answer:0.316 rad/s

Explanation:

Given

mass of Person m=65 kg

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Moment of inertia of Person I

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I=686.56 kg-m^2

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\omega _1=\frac{3.8}{3.25}=1.17 rad/s

Conserving Angular momentum

I\omega _1=(I+I_0)\omega _2 , where \omega _2=final\ angular\ velocity

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\omega _2=\frac{686.56}{2536.56}\times 1.17

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3 years ago
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