A square loop of side 20 cm and resistance 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. What should be the value of v₀ so that a steady current of 2 mA flows in the loop

Option 1 -

1 m/s

Option 2 -

10² m/s

Option 3 -

10^-2 cm/s

Option 4 -

1 cm/s

5 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$l = \frac{V_{0} B l}{R} = \frac{V_{0} \times 5 \times 20 \times 1 0^{- 2}}{4 + 1}$

$2 \times 1 0^{- 3} = V_{0} \times 20 \times 1 0^{- 2}$

$V_{0} = \frac{2 \times 1 0^{- 3}}{20 \times 1 0^{- 2}} = \frac{2}{2} \times 1 0^{- 2} = 1 \times 1 0^{- 2} m / s$

V₀ = 1 cm/s

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