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. The magnitude of the magnetic field (B) due to the loop ABCD at the origin (O) is: (a) (μ₀ I(b-a))/(24ab) (b) (μ₀ I)/(4π)[b-a/ab] (c) (μ₀ I)/(4π)[2(b-a)+π/3(a+b)] (d) zero

Answer: (a). AB and CD are radial, so they give no field at O. Arcs DA (radius a) and BC (radius b) carry current in opposite senses, θ = π/6. B = (μ₀ Iθ)/(4π)(1/a-1/b) = (μ₀ I)/(24)-a/ab So the answer is (a) (μ₀ I(b-a))/(24ab).

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

Q103JEE Main 2009 · AIEEEMedium

Biot-Savart Law & Circular Coils

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.

The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is:

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Answer: (a)

AB and CD are radial, so they give no field at O.
Arcs DA (radius $a$ ) and BC (radius $b$ ) carry current in opposite senses, $\theta = \dfrac{\pi}{6}$ .
$B = \dfrac{\mu_0 I\theta}{4\pi}\left(\dfrac{1}{a}-\dfrac{1}{b}\right) = \dfrac{\mu_0 I}{24}\cdot\dfrac{b-a}{ab}$
So the answer is (a) $\frac{\mu_0 I(b-a)}{24ab}$ .

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