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Physics Part-I

Electromagnetic Induction

Suppose the loop in Exercise 6.4 is stationary but the current

Physics Part-I

Class 12

NCERT

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has a magnitude of

5 \times 1 0^{-4} T

and the dip angle is 30°.

Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T

s^{-1}

. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm

s^{-1}

in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of

1 0^{-3} T c m^{-1}

along the negative x-direction (that is it increases by

1 0^{-3} T c m -1

as one moves in the negative x-direction), and it is decreasing in time at the rate of

1 0^{-3} T s^{-1}

Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

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Question

Medium

Solving time: 4 mins

Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T $s^{-1}$ . If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

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Physics Part-I (NCERT)

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s^{-1}

. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

14 mins ago

Discuss this question LIVE

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Text solutionVerified

A = 8 \times 2 = 16 c m^{2}

emf,

e = \frac{d ( A \times B )}{d t} = 0.32 \times 1 0^{- 4} V

Induced current,

i = e / R = 2 \times 1 0^{- 5} A

Power,

P = i^{2} R = 6.4 \times 1 0^{- 10} W

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5 \times 1 0^{-4} T

and the dip angle is 30°.

s^{-1}

. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm

s^{-1}

in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of

1 0^{-3} T c m^{-1}

along the negative x-direction (that is it increases by

1 0^{-3} T c m -1

as one moves in the negative x-direction), and it is decreasing in time at the rate of

1 0^{-3} T s^{-1}

Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

View all

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Question Text	Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T $s^{-1}$ . If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Topic	Electromagnetic Induction
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	76

Text solutionVerified

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