Question

A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed

with its plane perpendicular to the horizontal component of the earth’s

magnetic field. It is rotated about its vertical diameter through 180°

in 0.25 s. Estimate the magnitudes of the emf and current induced in

the coil. Horizontal component of the earth’s magnetic field at the

place is 3.0 × 10–5 T.​

Answer 1

Answer:

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Answer 2

The magnitudes of the emf and current induced in  the coil are $3.8\times10^3V$ and $1.9\times10^-^3A$ respectively.

Explanation:

The radius of the coil, r = 10 cm = 0.10 cm.

The number of the turns, N = 500.

The resistance of the coil, R = 2 ohm.

The time duration, dt = 0.25 s.

The earth’s magnetic field,  $B=3.0\times10^-^5 T$.

The emf induced in coil is given as

$emf=-N\frac{d\phi}{dt}$.

Here, $d\phi$ is the change in magnetic flux.

Initial magnetic flux through coil,

$\phi_i=BAcos0^0=3\times10^-^5T\times\pi(0.10cm)^2cos0$

$\phi_i=3\pi\times10^-^7Wb$

Final magnetic flux through coil,

$\phi_f=BAcos0^0=3\times10^-^5T\times\pi(0.10cm)^2cos180$

$\phi_f=-3\pi\times10^-^7Wb$.

so, emf

$emf=-500\times\frac{-3\pi\times10^-^7Wb-3\pi\times10^-^7Wb}{0.25s}$

$emf=500\times\frac{6\pi\times10^-^7Wb}{0.25s}$

$emf=3.8\times10^3V$

Therefore, induced current in the coil,

$i=\frac{emf}{R}=\frac{3.8\times10^3V}{2\Omega}$

$i = 1.9\times10^-^3A$

Thus, the magnitudes of the emf and current induced in  the coil are $3.8\times10^3V$ and $1.9\times10^-^3A$ respectively.

#Learn More: emf.

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