Q43 / 51 · JEE Advanced 2023 (Paper 2) · Physics

A rectangular conducting loop of length 4 cm and width 2 cm is in the xy-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction 32x∧+12y∧ with a constant speed v. The wire is carrying a steady current I=10 A in the positive x-direction. A current of 10 μA flows through the loop when it is at a distance d=4 cm from the wire. If the resistance of the loop is 0.1 Ω, then the value of v is _____ m s-1.

[Given: The permeability of free space μ0=4π×10-7 N A-2]

Figure

Correct answer: 4

Solution

The two sides of the rectangular loop which are perpendicular to the wire would not contribute in induced emf.

Angle of velocity vector with x-axis is tanα=1232=13. Therefore, α=30°.

For parallel sides:

E→=B→×v→

=μ0I2πx×v

Figure

⇒ Net emf =E1cos60o-E2cos60o×width

=12×2100×μ0Iv2π14100-18100

=1100×2×10-7×10×v×1008

=2.5v×10-7

Now, net EMF =i×R

⇒ v=10×10-6×0.12.5×10-7=4 m s-1

Hence, the correct answer is 4.

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