Question

An induced current of 2 A flows through a coil. The resistance of the coil is 10Ω . What is the change in magnetic flux associated with the coil in 1 ms ?

• A. 0.2 x10 ^-2 Wb
• B. 2 x10 ^-2 Wb
• C. 22 x10 ^-2 Wb
• D. 0.22 x10 ^-2 Wb

Answer 1

The Correct Option is B

Given:

• Induced current, \( I = 2 \, \text{A} \)
• Resistance of coil, \( R = 10 \, \Omega \)
• Time interval, \( \Delta t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \)
• Need to find: Change in magnetic flux \( \Delta \Phi \)

Step 1: Use Faraday’s Law of Electromagnetic Induction

\[ \text{Induced emf} \, (\mathcal{E}) = R \cdot I = 10 \times 2 = 20 \, \text{V} \]

Also, emf is related to change in flux by:

\[ \mathcal{E} = \frac{\Delta \Phi}{\Delta t} \Rightarrow \Delta \Phi = \mathcal{E} \cdot \Delta t = 20 \cdot 1 \times 10^{-3} = 20 \times 10^{-3} = 2 \times 10^{-2} \, \text{Wb} \]

The change in magnetic flux is \( {2 \times 10^{-2} \, \text{Wb}} \), so the correct answer is (B).

Answer 2

We use Faraday’s Law of Induction to solve this problem. The induced EMF in the coil can be expressed as: \[ \mathcal{E} = - \frac{\Delta \Phi}{\Delta t} \] Where: - \(\mathcal{E}\) is the induced EMF, - \(\Delta \Phi\) is the change in magnetic flux, - \(\Delta t\) is the time interval. Now, using Ohm's law, the induced EMF can also be expressed as: \[ \mathcal{E} = I \times R \] Where: - \(I = 2 \, A\) (induced current), - \(R = 10 \, \Omega\) (resistance of the coil). Thus, the induced EMF is: \[ \mathcal{E} = 2 \, \text{A} \times 10 \, \Omega = 20 \, \text{V} \] Since \(\mathcal{E} = - \frac{\Delta \Phi}{\Delta t}\), we can solve for the change in magnetic flux \(\Delta \Phi\): \[ \Delta \Phi = - \mathcal{E} \times \Delta t = - 20 \, \text{V} \times 1 \, \text{ms} = - 20 \times 10^{-3} \, \text{Wb} = - 2 \times 10^{-2} \, \text{Wb} \] Thus, the change in magnetic flux is \(2 \times 10^{-2} \, \text{Wb}\).