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A 1000-turn coil of wire 2.0 cm in diameter is in a magnetic field that drops from 0.10 T to 0 T in 10 ms. The axis of the coil is parallel to the field. What is the emf of the coil

Respuesta :

The Emf of the coil is 12.57

The Emf is represented by e.

e= N × dΦ / dt

where N is the number of turns, and Φ is the magnetic flux.

Φ= B.A

Φ = B × A × cos θ

where θ is the angle between field and area.

Since the area vector is parallel to the axis and also the field is parallel to the axis. Hence, the area vector is also parallel to the field.

Then, θ becomes zero.

cos (0)= 1

Hence, Φ= B ×A × 1

               = B × A

Therefore, the formula for Emf becomes:

e= N × d(B × A) / dt

Since A is constant

e  = N × A × d(B) / dt     .....(i)

Now, N = 1000

A = 2.0 cm

 = (2)² × 10⁻⁴

Putting values in equation (i)

e = 1000 × π × (2)² × 10⁻⁴ × (0.10 - 0 / 10) × 10³

e =12.57 V

Therefore, the emf of cell is 12.57

Read more about Emf of cell:

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