JEE Advanced 2021 · Paper 2 · Physics Q6

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PhysicsMultiple correct+4 / -2Question 6 of 57

Two concentric circular loops, one of radius $R$ and the other of radius $2R$ , lie in the $xy$ -plane with the origin as their common center, as shown in the figure. The smaller loop carries current $I_1$ in the anti-clockwise direction and the larger loop carries current $I_2$ in the clockwise direction, with $I_2 > 2I_1$ . $\vec B(x, y)$ denotes the magnetic field at a point $(x, y)$ in the $xy$ -plane. Which of the following statement(s) is(are) correct?

Two concentric circular loops centred at the origin O in the xy-plane: the inner of radius R carries I1 anticlockwise, the outer of radius 2R carries I2 clockwise.

\vec B(x, y)

is perpendicular to the

xy

-plane at any point in the plane

|\vec B(x, y)|

depends on

x

and

y

only through the radial distance

r = \sqrt{x^2 + y^2}

|\vec B(x, y)|

is non-zero at all points for

r < R

\vec B(x, y)

points normally outward from the

xy

-plane for all the points between the two loops

Answer and worked solution

Official answer: A, B

Symmetry. Both loops lie in the $xy$ -plane and are centred at the origin.

At any point in the plane, the field of each current element is perpendicular to the plane. (A) is true.
The arrangement looks the same after any rotation about the $z$ -axis, so $|\vec B|$ depends only on $r$ . (B) is true.

Inside, $r<R$ . At the centre,
$B=\frac{\mu_0I_1}{2R}-\frac{\mu_0I_2}{4R}=\frac{\mu_0}{4R}(2I_1-I_2)<0.$
But just inside the inner wire, the field of $I_1$ grows without bound (pointing out of the plane). So $B$ changes sign somewhere in $r<R$ and is zero there. (C) is false.

Between the loops. Just outside the inner wire, the field of $I_1$ points into the plane and grows without bound. So $\vec B$ is not outward everywhere between the loops. (D) is false.

(A), (B)

Topics:Biot-Savart law

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JEE Advanced 2021 · Paper 2, Physics question 6