Two concentric circular loops, one of radius R and the other of radius 2 R , lie in the xy-plane with the origin as their common centre, 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 clock wise direction, with I 2 2 I 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?

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

overrightarrow B ( x , y ) points normally outward from the xy-plane for all the points between the two loops

Physics · Moving Charges and Magnetic Field

JEE Advanced 2021 — Paper 2 — Question 6

Two concentric circular loops, one of radius R and the other of radius 2 R , lie in the xy-plane with the origin as their common centre, as shown in the figure.

The smaller loop carries current $\mathrm{I}_{1}$ in the anti-clockwise direction and the larger loop carries current $I_{2}$ in the clock wise direction, with

$I_{2}>2 I_{1}$ . $\vec{B}(x, y)$ denotes the magnetic field at a point ( $\mathrm{x}, \mathrm{y}$ ) in the xy-plane.

Which of the following statement(s) is(are) correct?

Option A:
$\vec{B}(x, y)$ is perpendicular to the xy-plane at any point in the plane
Correct
Option B:
$|\vec{B}(x, y)|$ depends on $x$ and $y$ only through the radial distance $r=\sqrt{x^{2}+y^{2}}$
Correct
Option C:
$|\vec{B}(x, y)|$ is non-zero at all points for $r<R$
Option D:
$\overrightarrow{\mathrm{B}}(\mathrm{x}, \mathrm{y})$ points normally outward from the xy-plane for all the points between the two loops

Answer: A, B

Step-by-step solution

Consider a circular loop of radius r in $\mathrm{x}-\mathrm{y}$ plane and having centre at origin

$\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=0$ $\mathrm{B} \oint \mathrm{d} \ell \cos \theta=0$

$\because B \neq 0 \quad$ for given $r$ $\Rightarrow \cos \theta=0$ $\theta=90^{\circ}$ Here $\mathrm{d} \ell$ is in xy plane

$\Rightarrow \mathrm{B}$ is normal to plane ( B can't be in xy plane as its magnetic lines would have been in radial direction)

Also, for given r, B must be same in magnitude for all points on loop of radius r. At centre $B=\left(\frac{\mu_{0} i_{1}}{2 R}-\frac{\mu_{0} i_{2}}{4 R}\right)$

(inwards) For point $P$ , Let field of inner loop increases $x_{1}$ times and that of outer loop increases $\mathrm{x}_{2}$ times

$\Rightarrow$ magnetic field at P $B_{P}=\left(x_{1} \frac{\mu_{0} i_{1}}{2 R}-x_{2} \frac{\mu_{0} i_{2}}{4 R}\right)$

For $B_{P}=0, i_{2}=\left(\frac{x_{1}}{x_{2}}\right) .\left(2 i_{1}\right)$ $\because B$ changes more rapidly as point P come closer to circumference.

$\Rightarrow x_{1}>x_{2}$ Or $\mathrm{i}_{2}>2 \mathrm{i}_{1}$ (which is given condition)

So, there are points inside inner loop where magnetic field will be zero.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law