Question

1. A strong electromagnet produces a uniform field of 1.60 T over a cross sectional area of 0.200 m2. A coil having 200 turns and a total resistance of 20.00 W is placed around the electromagnet. The current in the electromagnet is then smoothly decreased until it reaches zero in 20.0 ms. What is the current induced in the coil?

a. 0.160 A b. 0.800 A c. 160 A d.800 A

Answer 1

Answer:

A = 0.2 m2,

N = 200,

dB = 1.60 Wbm−2

dt= 20 ms 20/1000 sec

As E= dϕ/dt

=NA*dB/dt

= 200*0.2*1.60*1000/20

= 20*2*1.60*1000/20

= 2*1.60*1000

= 3.2*1000

= 3200 V

Current

I = V/R = 3200/20 = 160 A

Answer 2

Answer:

The current induced in the coil is 160 A i.e.option(c).

Explanation:

From the question we have,

The magnetic field(B)=1.60 T

The cross-sectional area(A)=0.200m²

The number of turns in the coil(N)=200

The total resistance of the coil(R)=20Ω

The time(t)=20 ms=20×10⁻³s

First, let's calculate emf(e),

$\mathrm{e=\frac{NAdB}{dt} }$              (1)

When we enter the values into equation (1), we obtain;

$\mathrm{e=\frac{200\times 0.2 \times 1.6}{20\times 10^{-3}} }$

$\mathrm{e=0.32\times 10^4}$

$\mathrm{e=3.2\times 10^4\ V}$         (2)

The current induced in the coil is calculated as,

$\mathrm{I=\frac{e}{R}}$              (3)

When we enter the values into equation (3), we obtain;

$\mathrm{I=\frac{3.2\times 10^3}{20}}$

$\mathrm{I=1.6\times 10^2\ A=160\ A}$

So, the current induced in the coil is 160 A i.e.option(c).

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