AIIMS2007Physics-Electromagnetic Induction

AIIMS 2007 Physics Mutual Inductance MCQ Question

Type: MCQ-conceptual-Medium-Class 12

With the decrease of current in the primary coil from 2 amperes to zero value in 0.01s the emf generated in the secondary coil is 1000 volts. The mutual inductance of the two coils is

A

1.25 H

B

2.50 H

C

5.00 H

D

10.00 H

Correct Answer

Option C

Detailed Explanation

To solve the problem of finding the mutual inductance of two coils when there is a change in current in the primary coil, we can use the formula that relates the induced electromotive force (emf) in the secondary coil to the mutual inductance and the rate of change of current in the primary coil.

Relevant Formula

The induced emf (E\mathcal{E}) in the secondary coil due to a change in current in the primary coil is given by:

E=MdIdt\mathcal{E} = -M \frac{dI}{dt}

where:

  • E\mathcal{E} is the induced emf in volts,
  • MM is the mutual inductance in henries (H),
  • dIdI is the change in current in amperes,
  • dtdt is the change in time in seconds.

Given Data

From the problem:

  • The initial current Ii=2I_i = 2 A,
  • The final current If=0I_f = 0 A,
  • The time interval dt=0.01dt = 0.01 s,
  • The induced emf E=1000\mathcal{E} = 1000 V.

Step 1: Calculate the Change in Current

The change in current dIdI can be calculated as:

dI=IfIi=02=2 AdI = I_f - I_i = 0 - 2 = -2 \text{ A}

Step 2: Substitute Values into the Formula

Now we substitute the values into the formula for induced emf:

1000=M20.011000 = -M \frac{-2}{0.01}

Step 3: Simplifying the Equation

Rearranging the equation gives:

1000=M20.011000 = M \cdot \frac{2}{0.01}

Calculating 20.01\frac{2}{0.01}:

20.01=200\frac{2}{0.01} = 200

So, we can rewrite our equation as:

1000=M2001000 = M \cdot 200

Step 4: Solve for Mutual Inductance MM

Now, divide both sides by 200:

M=1000200=5 HM = \frac{1000}{200} = 5 \text{ H}

Conclusion

Thus, the mutual inductance MM of the two coils is 55 H. Therefore, the correct answer is:

C) 5.00 H

Why Other Options are Incorrect

  • A) 1.25 H: This value would imply a significantly lower rate of change of current than what was calculated, hence not matching the given induced emf.
  • B) 2.50 H: Similar reasoning applies; this would not produce the required induced emf with the given change in current.
  • D) 10.00 H: This value would produce a much higher induced emf than 1000 V for the given change in current, making it incorrect.

By thoroughly analyzing the problem using the appropriate formula and calculations, we confirm that the mutual inductance of the two coils is indeed 55 H, validating option C as the correct answer.

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