JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 2)
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?
_1st_September_Evening_Shift_en_2_1.png)
1 m/s
1 cm/s
102 m/s
10$$-$$2 cm/s
Explanation
According to given circuit diagram, equivalent resistance between point P and Q.
$${R_{PQ}} = (4 + 4)||(4 + 4)$$
$$ = {{8 \times 8} \over {8 + 8}} = 4\,\Omega $$
The equivalent circuit can be drawn as,
_1st_September_Evening_Shift_en_2_2.png)
Equivalent resistance, Req = 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 $$\times$$ 10-3 A
Induced emf, e = Bv0I
Induced current, I = e / Req
Substituting the values in the above equation, we get
$$2 \times {10^{ - 3}} = {{5 \times {v_0} \times 0.2} \over 5}$$
$$\Rightarrow$$ v0 = 10-2 m/s = 1 cm/s
$$\therefore$$ The value of v0 = 1 cm/s, so that a steady current of 2 mA flows in the loop.
$${R_{PQ}} = (4 + 4)||(4 + 4)$$
$$ = {{8 \times 8} \over {8 + 8}} = 4\,\Omega $$
The equivalent circuit can be drawn as,
Equivalent resistance, Req = 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 $$\times$$ 10-3 A
Induced emf, e = Bv0I
Induced current, I = e / Req
Substituting the values in the above equation, we get
$$2 \times {10^{ - 3}} = {{5 \times {v_0} \times 0.2} \over 5}$$
$$\Rightarrow$$ v0 = 10-2 m/s = 1 cm/s
$$\therefore$$ The value of v0 = 1 cm/s, so that a steady current of 2 mA flows in the loop.
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