Question 76: Electromagnetic Induction

Question

A square loop of side 20 cm and resistance 1$\Omega$ is moved towards right with a constant speed v
0
. The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4$\Omega$. What should be the value of v
0
so that a steady current of 2 mA flows in the loop?
![](https://app-content.cdn.examgoal.net/fly/@width/image/1kx9suidr/12dbe242-2546-48c8-83d1-a9941db95493/d5c191f0-5ee4-11ec-9c4b-9f471c110c87/file-1kx9suids.png?format=png)

Accepted Answer

According to given circuit diagram, equivalent resistance between point P and Q.
${R_{PQ}} = (4 + 4)||(4 + 4)$
$ = {{8 	imes 8} \over {8 + 8}} = 4\,\Omega $
The equivalent circuit can be drawn as,
Equivalent resistance, R
eq
= 4 + 1 = 5 $\Omega$
Magnetic field, B = 5T
The side of the square loop, I = 20 cm = 0.20 m
The steady value of the current, I = 2 mA = 2 $	imes$ 10
-3
A
Induced emf, e = Bv
0
I
Induced current, I = e / R
eq
Substituting the values in the above equation, we get
$2 	imes {10^{ - 3}} = {{5 	imes {v_0} 	imes 0.2} \over 5}$
$\Rightarrow$ v
0
= 10
-2
m/s = 1 cm/s
$	herefore$ The value of v
0
= 1 cm/s, so that a steady current of 2 mA flows in the loop.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1kx9wir0q/8ef16fdb-1a5f-4e75-bc68-09ca53231c0d/333c5aa0-5ef3-11ec-85b8-4b78fdd0cc43/file-1kx9wir0r.png?format=png)

Answer

1 m/s

Answer

1 cm/s

Answer

10
2
m/s

Answer

10
$-$2
cm/s

Electromagnetic Induction

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