A generator can be made using the component of Earth’s magnetic field that is parallel to Earth’s surface. A 140-turn square wire coil with an area of 4.30 ×10−2 m2 is mounted on a shaft so that the cross-sectional area of the coil is perpendicular to the ground. The shaft then rotates with a frequency of 21.4 Hz. The horizontal component of the Earth’s magnetic field at the location of the loop is 3.00 ×10−5 T.

Calculate the maximum emf induced in the coil by Earth's magnetic field.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-05-15T02:59:27+02:00

Answer:

The maximum induced emf is [tex]\epsilon = 0.024 V[/tex]

Explanation:

From the question we are told that

The number of turns is [tex]N = 140 \ turns[/tex]

The area of the square wire coil is [tex]A = 4.30 * 10^{-2}m^2[/tex]

The frequency of rotation is [tex]f = 21.4 Hz[/tex]

The earth magnetic field at that location is [tex]B = 3.00 *10^{-5} T[/tex]

The induced emf is mathematically represented as

[tex]\epsilon = N \ B \ A \ w sin(wt)[/tex]

At the maximum

[tex]sin (wt) = 1[/tex]

So

[tex]\epsilon = N \ B \ A \ w[/tex]

Where

[tex]w[/tex] is the angular velocity which is mathematically represented as

[tex]w =2 \pi f[/tex]

Substituting values

[tex]w = 2 * 3.142 * 21.4[/tex]

[tex]w = 134.5 rad/sec[/tex]

The maximum induced emf is

[tex]\epsilon = 140 * 3.00*10^{-5} * (4.30 *10^{-2}) * (134.5)[/tex]

[tex]\epsilon = 0.024 V[/tex]