A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30°. Another straight thin wire steady current I₁ flowing out of the plane of the paper is kept at the origin. Due to the presence of the current I₁ at the origin: (a) The forces on AD are BC are zero. (b) The magnitude of the net force on the loop is given by (I₁ I)/(4π)μ₀[2(b-a) + π/3(a+b)]. (c) The magnitude of the net force on the loop is given by (μ₀ I I₁)/(24ab)(b-a). (d) The forces on AB and DC are zero.

Answer: (a). Field of I₁ forms circles centred at O, so it is tangent to arcs AD and BC. Along the arcs d l ∥ B, so d l × B = 0: forces on AD and BC are zero. AB and CD are radial, so d l ⊥ B; each gets a force (μ₀ I I₁)/(2π)/a perpendicular to the paper, in opposite directions (net force zero, but a torque). So the answer is (a).

A current loop ABCD is held fixed on the plane of the paper as shown… | JEE Main 2009 (AIEEE)

Q64JEE Main 2009 · AIEEEDifficult

Force on Current-Carrying Conductors

A current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$ ) and $DA$ (radius $= a$ ) of the loop are joined by two straight wires $AB$ and $CD$ . A steady current $I$ is flowing in the loop. Angle made by $AB$ and $CD$ at the origin $O$ is $30^\circ.$ Another straight thin wire steady current $I_1$ flowing out of the plane of the paper is kept at the origin.

Due to the presence of the current $I_1$ at the origin:

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Answer: (a)

Field of $I_1$ forms circles centred at O, so it is tangent to arcs AD and BC.
Along the arcs $d\vec l \parallel \vec B$ , so $d\vec l \times \vec B = 0$ : forces on AD and BC are zero.
AB and CD are radial, so $d\vec l \perp \vec B$ ; each gets a force $\frac{\mu_0 I I_1}{2\pi}\ln\frac{b}{a}$ perpendicular to the paper, in opposite directions (net force zero, but a torque).
So the answer is (a).

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Force on Current-Carrying Conductors question from JEE Main 2009 (AIEEE)

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