Question

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum emf generated is:

Answer 1

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For maximum emf, $\sin\omega t=1$

So,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

So, the maximum emf generated in the loop is 5.32 V.

Answer 2

The maximum emf generates in the conducting loop is 5.32 V.

Given data:

The number of turns of loop is, n = 10.

The radius of loops is, r = 3.0 cm = 0.03 m.

The number of revolutions per second is, f = 60.

The strength of magnetic field is, B = 0.50 T.

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The expression for the emf generated in the conducting loop is given as,

$e = nBA\times sin \omega t$

here, A is the area of loop and its value is, $A =\pi r^{2}$.

For maximum emf, the value of $sin \omega t$ should be maximum. So, $sin \omega t=\omega$. Then calculating the maximum emf as,

$e_{max.} = nBA \times \omega\\\\e_{max.} = nB \times (\pi \times r^{2}) \times (2 \pi f)\\\\e_{max.} = 10 \times 0.50 \times (\pi \times (0.03)^{2}) \times (2 \pi \times 60)\\\\e_{max.}=5.32 \;\rm V$

Thus, we can conclude that the maximum emf generates in the conducting loop is 5.32 V.

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