Question 91: Electromagnetic Induction

Question

A square loop of side 20 cm and resistance 1$$\Omega$$ is moved towards right with a constant speed v0. 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 v0 so that a steady current of 2 mA flows in the loop?

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.

Answer

1 m/s

Answer

1 cm/s

Answer

102 m/s

Answer

10$$-$$2 cm/s

Electromagnetic Induction

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