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nikdorinn [45]
3 years ago
10

An elastic conducting material is stretched into a circular loop of 10.7 cm radius. It is placed with its plane perpendicular to

a uniform 0.803 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 81.0 cm/s. What emf is induced in volts in the loop at that instant?
Physics
1 answer:
valina [46]3 years ago
3 0

Answer:

0.44 V

Explanation:

Parameters given:

Radius of loop, r = 10.7 cm = 0.107 m

Magnetic field strength, B = 0.803 T

Rate of shrinkage of Radius, dr/dt = 81 cm/s = 0.81m/s

EMF induced is given in terms of magnetic Flux, Φ, as:

EMF = dΦ/dt

Magnetic Flux, Φ, is given as:

Φ = B * A (where A is area of loop)

Therefore, EMF is:

EMF = d(B*A)/dt

The area of the loop is given as A = πr²

EMF = d(Bπr²) / dt = Bπ*d(r²)/dt

=> EMF = Bπ*2r*(dr/dt)

EMF = 0.803 * π * 2 * 0.107 * 0.81

EMF = 0.44 V

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Answer:

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

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For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

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S = 6.956 m

S = 7.0 m

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