Question

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 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 $I_2>2I_1,\vec{B}(x,y)$ denotes the magnetic field at a point $(x,y)$ in the $xy$-plane. Which of the following statement (s) is ( are ) correct?

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

Answer 1

Answer: (A), (B)

Consider a circular loop of radius $r$ in $x-y$ plane and having centre at origin
$\oint \vec{B}\cdot \overrightarrow{d\ell }=0$
$B\oint d\ell \cos \theta =0$
$\because B\ne 0$ for given $r$
$\Rightarrow \cos \theta =0$
$\theta =90^{\circ }$

Here $d\ell$ is in $xy$ plane $\Rightarrow B$ 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 $B=\left(\frac{{\mu }_0{i}_1}{2R}-\frac{{\mu }_0{i}_2}{4R}\right)$
(inwards)

For point $P$, Let field of inner loop increases $x_1$ times and that of outer loop increases $x_2$ times
$\Rightarrow$ magnetic field at $P$
$B_P=\left(x_1\frac{{\mu }_0{i}_1}{2R}-x_2\frac{{\mu }_0{i}_2}{4R}\right)$
For $B_P=0,i_2=\left(\frac{x_1}{x_2}\right)\cdot \left(2i_1\right)$
$\because B$ changes more rapidly as point $P$ come closer to circumference.
$\Rightarrow x_1>x_2$
Or $i_2>2i_1$ (which is given condition)
So, there are points inside inner loop where magnetic field will be zero.