Question

The axis of a 100 turn circular coil (area of cross-seciton 3.85 × 10–3 m2) is parallel to a
uniform magnetic field. The magnitude of the field changes at a constant range from 25 mT
to 50 mT. in 250 ms. Calculate the magnitude of induced emf across the coil.

Answer 1

The induced emf will be 0.0385V

Explanation:

In the question the following data is given  

Number of tuns = N = 100 turns

Area of coil = A = 3.85 × 10–3 m2

dB = 50-25 = 25mT

dt = 250 ms

We know that:

E = N.dΦ/dt

E = (NA)dB/dt

E = 100 × (25m) × (3.85 × 10–3) / 250m

E = 0.0385 V

Answer 2

Given that,

Number of turns in the coil are 100

Area of cross section of the coil is $3.85\times 10^{-3}\ m^2$

Initial magnetic field is 25 mT and final magnetic field is 50 mT

Time of duration of changing magnetic field is 250 ms.

To find,

The magnitude of induced emf across the coil.

Solution,

According to Faraday's law, the induced emf is given by :

$\epsilon=-\dfrac{Nd\phi}{dt}\\\\\epsilon=-\dfrac{d(NBA)}{dt}\\\\\epsilon=-NA\dfrac{dB}{dt}\\\\\epsilon=-NA\dfrac{B_f-B_i}{dt}\\\\\epsilon=-100\times 3.85\times 10^{-3}\times \dfrac{(50-25)\times 10^{-3}}{250\times 10^{-3}}\\\\\epsilon=0.0385\ V$

So, the magnitude of induced emf across the coil is 0.0385 volts.

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Induced emf

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