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A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0215 W/m2W/m2, and the wavelength of the wave is 6.90 mm.What is the maximum emf induced in the loop?

Express your answer with the appropriate units.

1 Answer

3 votes

Answer:

The induced emf is
\epsilon = 0.1041 \ V

Step-by-step explanation:

From the question we are told that

The radius of the circular loop is
r = 9.50 \ cm = 0.095 \ m

The intensity of the wave is
I = 0.0215 \ W/m^2

The wavelength is
\lambda = 6.90\ m

Generally the intensity is mathematically represented as


I = ( c * B^2 )/( 2 * \mu_o )

Here
\mu_o is the permeability of free space with value


\mu_o = 4 \pi *10^(-7) N/A^2

B is the magnetic field which can be mathematically represented from the equation as


B = \sqrt{ ( 2 * \mu_o * I )/( c) }

substituting values


B = \sqrt{ ( 2 * 4\pi *10^(-7) * 0.0215 )/( 3.0*10^(8)) }


B = 1.342 *10^(-8) \ T

The area is mathematically represented as


A = \pi r^2

substituting values


A = 3.142 * (0.095)^2


A = 0.0284

The angular velocity is mathematically represented as


w = 2 * \pi * (c)/(\lambda )

substituting values


w = 2 * 3.142 * (3.0*10^(8))/( 6.90 )


w = 2.732 *10^(8) rad \ s^(-1)

Generally the induced emf is mathematically represented as


\epsilon = N * B * A * w * sin (wt )

At maximum induced emf
sin (wt) = 1

So


\epsilon = N * B * A * w

substituting values


\epsilon = 1 * 1.342 *10^(-8) * 0.0284 *2.732 *10^(8)


\epsilon = 0.1041 \ V

User Arth Tilva
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