A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from +ve z direction and lying in x y plane is shown in figure. The plot of \hatj component of magnetic field (By) at a distance ' a ' (less than radius of the coil) and on y z plane vs z coordinate looks like

Magnetics PYQ — JEE Main 2023 · 29 Jan · Shift 1 · Q61 with Solution

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Correct answer: D

Understand the loop and current direction

A circular loop lies in the $xy$ -plane, centered at the origin. Current is anticlockwise as seen from $+z$ .

By the right-hand rule, the magnetic moment is along $+z$ .

We are asked about the $\hat{j}$ -component of magnetic field, i.e. $B_y$ , at points in the $yz$ -plane with fixed $y=a$ and varying $z$ , where $a<R$ (inside the projection of the loop).

So the observation point is

P=(0,a,z).

Use symmetry to infer the direction of field components

Because the loop is symmetric about the $yz$ -plane, at points on the $yz$ -plane the $x$ -component cancels:

B_x=0.

So only $B_y$ and $B_z$ may exist.

We need the variation of $B_y$ with $z$ .

Determine the sign of $B_y$ at $z=0$

At the point $(0,a,0)$ in the plane of the loop and inside the loop, the magnetic field lines return opposite to the magnetic moment direction outside the conductor, and near the wire plane the in-plane component at such a point is along $-\hat y$ .

We can verify this more formally from Biot–Savart law.

Take an element of the loop at angle $\phi$ :

\vec r\,'=(R\cos\phi, R\sin\phi,0),

with current element

d\vec l = (-R\sin\phi, R\cos\phi,0)\,d\phi.

For point $P=(0,a,z)$ ,

\vec R = \vec r-\vec r\,' = (-R\cos\phi, a-R\sin\phi, z).

Then

d\vec B \propto d\vec l \times \vec R.

Its $y$ -component is

(d\vec l \times \vec R)_y = zR\sin\phi\, d\phi.

Hence

B_y \propto \int_0^{2\pi} \frac{z\sin\phi\,d\phi}{\left(R^2+a^2+z^2-2aR\sin\phi\right)^{3/2}}.

This immediately shows:

$B_y=0$ at $z=0$ ,
$B_y$ is an odd function of $z$ .

So the graph must pass through the origin and satisfy

B_y(-z)=-B_y(z).

Find the sign for $z>0$

For small positive $z$ , the sign of $B_y$ is the same as the sign of

\int_0^{2\pi} \frac{\sin\phi\,d\phi}{\left(R^2+a^2+z^2-2aR\sin\phi\right)^{3/2}}.

When $\sin\phi>0$ (upper half of loop), the denominator is smaller because of $-2aR\sin\phi$ , so those positive contributions dominate over the negative ones from the lower half.

Therefore for $z>0$ ,

B_y>0.

And for $z<0$ ,

B_y<0.

Thus the curve is an odd curve crossing origin with positive slope near $z=0$ .

Behavior for large $|z|$

Far from the loop, field behaves like that of a magnetic dipole along $+z$ . Away from the axis, the transverse component $B_y$ tends to zero as $|z|\to\infty$ .

So the graph must:

be negative for large negative $z$ but approach $0$ ,
pass through $0$ at $z=0$ ,
become positive for positive $z$ ,
again approach $0$ as $z\to +\infty$ .

So it is a two-lobed odd curve.

Match with the options

The correct plot is the one that is:

odd in $z$ ,
passes through origin,
has $B_y<0$ for $z<0$ ,
has $B_y>0$ for $z>0$ ,
tends to $0$ for large $|z|$ .

This corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Derived answer: D

So they agree.

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View written solutionFree

Correct answer: D

Understand the loop and current direction

A circular loop lies in the $xy$ -plane, centered at the origin. Current is anticlockwise as seen from $+z$ .

By the right-hand rule, the magnetic moment is along $+z$ .

We are asked about the $\hat{j}$ -component of magnetic field, i.e. $B_y$ , at points in the $yz$ -plane with fixed $y=a$ and varying $z$ , where $a<R$ (inside the projection of the loop).

So the observation point is

P=(0,a,z).

Use symmetry to infer the direction of field components

Because the loop is symmetric about the $yz$ -plane, at points on the $yz$ -plane the $x$ -component cancels:

B_x=0.

So only $B_y$ and $B_z$ may exist.

We need the variation of $B_y$ with $z$ .

Determine the sign of $B_y$ at $z=0$

We can verify this more formally from Biot–Savart law.

Take an element of the loop at angle $\phi$ :

\vec r\,'=(R\cos\phi, R\sin\phi,0),

with current element

d\vec l = (-R\sin\phi, R\cos\phi,0)\,d\phi.

For point $P=(0,a,z)$ ,

\vec R = \vec r-\vec r\,' = (-R\cos\phi, a-R\sin\phi, z).

Then

d\vec B \propto d\vec l \times \vec R.

Its $y$ -component is

(d\vec l \times \vec R)_y = zR\sin\phi\, d\phi.

Hence

B_y \propto \int_0^{2\pi} \frac{z\sin\phi\,d\phi}{\left(R^2+a^2+z^2-2aR\sin\phi\right)^{3/2}}.

This immediately shows:

$B_y=0$ at $z=0$ ,
$B_y$ is an odd function of $z$ .

So the graph must pass through the origin and satisfy

B_y(-z)=-B_y(z).

Find the sign for $z>0$

For small positive $z$ , the sign of $B_y$ is the same as the sign of

\int_0^{2\pi} \frac{\sin\phi\,d\phi}{\left(R^2+a^2+z^2-2aR\sin\phi\right)^{3/2}}.

When $\sin\phi>0$ (upper half of loop), the denominator is smaller because of $-2aR\sin\phi$ , so those positive contributions dominate over the negative ones from the lower half.

Therefore for $z>0$ ,

B_y>0.

And for $z<0$ ,

B_y<0.

Thus the curve is an odd curve crossing origin with positive slope near $z=0$ .

Behavior for large $|z|$

Far from the loop, field behaves like that of a magnetic dipole along $+z$ . Away from the axis, the transverse component $B_y$ tends to zero as $|z|\to\infty$ .

So the graph must:

be negative for large negative $z$ but approach $0$ ,
pass through $0$ at $z=0$ ,
become positive for positive $z$ ,
again approach $0$ as $z\to +\infty$ .

So it is a two-lobed odd curve.

Match with the options

The correct plot is the one that is:

odd in $z$ ,
passes through origin,
has $B_y<0$ for $z<0$ ,
has $B_y>0$ for $z>0$ ,
tends to $0$ for large $|z|$ .

This corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Derived answer: D

So they agree.

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