A circular wire loop 53 cm in diameter has resistance 120 and lies in a horizont

Question

A circular wire loop

53 cm in diameter has resistance 120 and lies in a horizontal plane. A uniform magnetic field points vertically downward, and in 25 ms it increases linearly from 5.0 mT to 55 mT.

Part A

Find the magnetic flux through the loop at the beginning.

Part B

Find the magnetic flux through the loop at the end of the 25-ms period.

Part C

What's the loop current during this time?

Express your answer with the appropriate units.

Part D

Which way does this current flow? Select the correct answer and explanation.

Which way does this current flow? Select the correct answer and explanation.

I is counterclockwise when viewed from above the loop because induced field is upward. I is clockwise when viewed from above the loop because induced field is downward. I is clockwise when viewed from above the loop because induced field is upward. I is counterclockwise when viewed from above the loop because induced field is downward

Explanation / Answer

radius of the loop is r = 53cm/2 = 26.5cm = 0.265m

Resistance of the wire is R = 120

Let us suppose the field was B1 = 5mT at t = 0

At t = 25ms the field becomes  B2 = 55mT

Part A

at the beginning, i.e., at t = 0 field is B1 = 5mT.

Area of the loop is A = r2 = 3.14(0.265m)2 = 0.22m2

Now let us suppose that the positive normal to the area A is downward. Thus, angle between the field B1 and normal to the area A is 00.

Now flux is given by:

= B.A = BAcos00 = BA = (5mT)0.22m2 = 1.1X10-3T-m2

Part B

At the end of 25ms, becomes  B2 = 55mT

So, = B.A = BAcos00 = BA = (55mT)0.22m2 = 12.1X10-3T-m2

Part C

Since B varies lineraly from t = 0 to t = 25ms from B1 = 5mT to B2 = 55mT, B depends on t as per the following eauation

B = 2t + 5mT

where B and t are in mT and ms respectively.

So the flux through the coil during this time is variable and it given by:

= B.A = (2t + 5mT).(0.22m2)

differentiating both sides with respect to time we get:

d/dt = (2).(0.22) = 0.44

Now emf induced in the coil is E = -d/dt = -0.44V

So, the magnitude of the current through the coil is I = E/R = (0.44V)/(120) = 3.66X10-3A

Part D

The magnetic field is downwards and increasing and hence the flux is increasing. So as per the Lenz's law the current would be induced in a direction such that it weakens the increasing flux. So the direction of current would be counterclockwise as viewed from above the coil and thus the direction of induced magnetic field will be upwards.

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige.....

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