A current carrying circular loop of radius ' R ' and current carrying… | MHT CET 2024 (11 May, Shift 2)

Q32MHT CET 2024 · 11 MAY · SHIFT 2Medium

Magnetic Field of Straight Wires

A current carrying circular loop of radius ' $R$ ' and current carrying long straight wire are placed in the same plane. The current through circular loop and long straight wire are ' $I_c$ ' and ' $I_w$ ' respectively. The perpendicular distance between centre of the circular loop and wire is ' d '. The magnetic field at the centre of the loop will be zero when separation ' $d$ ' is equal to

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Answer: (a)

Loop centre: $B_c = \frac{\mu_0 I_c}{2R}$ ; wire: $B_w = \frac{\mu_0 I_w}{2\pi d}$ . Net zero when they are equal and opposite.
$\frac{\mu_0 I_c}{2R} = \frac{\mu_0 I_w}{2\pi d}$
$d = \frac{R I_w}{\pi I_c}$
So the answer is (a).

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