Question

A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad s^-1. If the vertical component of earth’s magnetic field at that place is 2 × 10^-5 T and electrical resistance of the coil is 12.56 Ω, then the maximum induced current in the coil will be:

• A. 0.25 A
• B. 1.5 A
• C. 1 A
• D. 2 A

Answer 1

The Correct Option is C

To find the maximum induced current in the coil, we use Faraday's law of electromagnetic induction, which states the induced electromotive force (emf) in a coil with N turns is given by:

\[ \text{emf} = -N \frac{d\Phi}{dt} \]

where \(\Phi\) is the magnetic flux. The flux \(\Phi\) through the coil is given by:

\[ \Phi = B \cdot A \cdot \cos(\theta) \]

where \(B\) is the magnetic field, \(A\) is the area of the coil, and \(\theta\) is the angle between the magnetic field and normal to the coil's plane.

The coil's area \(A\) can be calculated as:

\[ A = \pi r^2 = \pi \times (10)^2 = 100\pi \text{ m}^2 \]

Given the vertical component of Earth's magnetic field is \(2 \times 10^{-5} \text{ T}\), the coil rotates at an angular velocity \(\omega = 2 \text{ rad/s}\). We find:

\[ \theta = \omega t \] \[ \cos(\theta) = \cos(2t) \]

Hence, the flux becomes:

\[ \Phi = B \cdot A \cdot \cos(\omega t) = 2 \times 10^{-5} \times 100\pi \cdot \cos(2t) \]

Then, the emf is the derivative of flux:

\[ \text{emf} = -N \frac{d\Phi}{dt} \] \[ = -1000 \cdot \frac{d}{dt}(2 \times 10^{-3} \pi \cdot \cos(2t)) \]

Using the chain rule, we have:

\[ \frac{d}{dt}(\cos(2t)) = -2\sin(2t) \]

Thus,

\[ \text{emf} = -1000 \times 2 \times 10^{-3} \pi \times (-2\sin(2t)) \] \[ = 4 \times 10^{-3} \pi \times 1000 \cdot \sin(2t) \] \[ = 4\pi \cdot \sin(2t) \text{ V} \]

The maximum emf is when \(\sin(2t) = 1\), leading to:

\[ \text{Maximum emf} = 4\pi \text{ V}\]

Ohm's Law relates current \(I\) to emf (\(V\)) via resistance \(R\):

\[ I = \frac{\text{emf}}{R} \]

Where the resistance of the coil is 12.56 Ω. Thus, the maximum current is:

\[ I_{\text{max}} = \frac{4\pi}{12.56} \text{ A} \] \[ = 1 \text{ A}\]

Therefore, the maximum induced current in the coil is 1 A.

Answer 2

E_max = NWAB
E_max = \(1000 \times 2 \times \pi \times (10)^2 \times 2 \times (10)^{-5}\)
E_max = \(12.56\) V
I_max = \(\frac{E_{max}}{R}\)

I_max = \(\frac{12.56}{12.56}\)
I_max = \(1\) A

So, the correct option is (C): 1 A