A square loop of side 2 a and carrying current I is kept in xz plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be :

2\mu _0I^2a^2b \over \pi ( a^2 + b^2 )

\mu _0I^2a^2b \over 2\pi ( a^2 + b^2 )

Magnetics PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q43 with Solution

View written solutionFree

Correct answer: A

Geometry and magnetic field due to the long wire

A long straight wire is parallel to the $z$ -axis and passes through $(0,b,0)$ .
The square loop lies in the $xz$ -plane ( $y=0$ ), centered at origin, with side $2a$ .

So the loop spans:

$x = \pm a$
$z = \pm a$
$y=0$

At a point $(x,0,z)$ on the loop, the perpendicular distance from the wire is

r=\sqrt{x^2+b^2}.

Magnetic field magnitude due to the long wire is

B=\frac{\mu_0 I}{2\pi r}=\frac{\mu_0 I}{2\pi\sqrt{x^2+b^2}}.

Using right-hand rule, at points on the plane $y=0$ , the field lies in the $xy$ -plane. Its $y$ -component is

B_y=\frac{\mu_0 I}{2\pi}\frac{x}{x^2+b^2}.

(The $x$ -component does not contribute to torque about $z$ for the horizontal sides, and the vertical sides give cancelling contributions about $z$ .)

Force on loop elements

Use

d\vec F=I\,d\vec l\times \vec B.

We need torque about the $z$ -axis:

\tau_z=(\vec r\times d\vec F)_z.

Vertical sides ( $x=\pm a$ )

Here $d\vec l=\pm dz\,\hat k$ .

Then

d\vec F=I(\pm dz\,\hat k)\times \vec B.

These forces lie in the $xy$ -plane. On integrating over the two opposite vertical sides, the torques about the $z$ -axis cancel by symmetry.

So net contribution from vertical sides to torque about $z$ is

\tau_z^{(vertical)}=0.

Top and bottom sides

These are the sides at $z=+a$ and $z=-a$ , with current along $\pm \hat i$ .

For an element on these sides,

d\vec l=\pm dx\,\hat i.

Then

d\vec F=I(\pm dx\,\hat i)\times \vec B.

Only the $B_y$ component contributes, giving force along $\pm \hat k$ :

dF_z=I B_y\,dx

for the top side and same-sign contribution to torque from the bottom side as well.

Now torque about $z$ from a vertical force element is

d\tau_z = x\,dF_y - y\,dF_x = 0

if force were only along $z$ . So instead it is easier to use the magnetic dipole approximation since $b\gg a$ .

Dipole moment of the square loop

Area of loop:

A=(2a)^2=4a^2.

Magnetic dipole moment magnitude:

m=IA=4Ia^2.

Since the loop lies in the $xz$ -plane, its normal is along $\pm \hat y$ . Thus

\vec m = 4Ia^2\,\hat y

(in magnitude; sign only affects direction, not magnitude of torque).

Magnetic field at the centre of the loop

Since $b\gg a$ , field is approximately uniform over the loop and equal to field at the origin.

Distance of origin from the wire is $b$ , so

B=\frac{\mu_0 I}{2\pi b}.

Direction at origin is along $\hat x$ (or $-\hat x$ depending on current direction), perpendicular to $\vec m$ . Hence angle between $\vec m$ and $\vec B$ is $90^\circ$ .

Torque on magnetic dipole

Magnitude of torque is

\tau = mB\sin 90^\circ = mB.

So,

\tau = (4Ia^2)\left(\frac{\mu_0 I}{2\pi b}\right) =\frac{2\mu_0 I^2 a^2}{\pi b}.

Match with options

\boxed{\tau=\frac{2\mu_0 I^2 a^2}{\pi b}}

This corresponds to Option A.

Comparison with stored answer

Stored correct answer: A
Derived answer: A

So they agree.

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View written solutionFree

Correct answer: A

Geometry and magnetic field due to the long wire

A long straight wire is parallel to the $z$ -axis and passes through $(0,b,0)$ .
The square loop lies in the $xz$ -plane ( $y=0$ ), centered at origin, with side $2a$ .

So the loop spans:

$x = \pm a$
$z = \pm a$
$y=0$

At a point $(x,0,z)$ on the loop, the perpendicular distance from the wire is

r=\sqrt{x^2+b^2}.

Magnetic field magnitude due to the long wire is

B=\frac{\mu_0 I}{2\pi r}=\frac{\mu_0 I}{2\pi\sqrt{x^2+b^2}}.

Using right-hand rule, at points on the plane $y=0$ , the field lies in the $xy$ -plane. Its $y$ -component is

B_y=\frac{\mu_0 I}{2\pi}\frac{x}{x^2+b^2}.

(The $x$ -component does not contribute to torque about $z$ for the horizontal sides, and the vertical sides give cancelling contributions about $z$ .)

Force on loop elements

Use

d\vec F=I\,d\vec l\times \vec B.

We need torque about the $z$ -axis:

\tau_z=(\vec r\times d\vec F)_z.

Vertical sides ( $x=\pm a$ )

Here $d\vec l=\pm dz\,\hat k$ .

Then

d\vec F=I(\pm dz\,\hat k)\times \vec B.

These forces lie in the $xy$ -plane. On integrating over the two opposite vertical sides, the torques about the $z$ -axis cancel by symmetry.

So net contribution from vertical sides to torque about $z$ is

\tau_z^{(vertical)}=0.

Top and bottom sides

These are the sides at $z=+a$ and $z=-a$ , with current along $\pm \hat i$ .

For an element on these sides,

d\vec l=\pm dx\,\hat i.

Then

d\vec F=I(\pm dx\,\hat i)\times \vec B.

Only the $B_y$ component contributes, giving force along $\pm \hat k$ :

dF_z=I B_y\,dx

for the top side and same-sign contribution to torque from the bottom side as well.

Now torque about $z$ from a vertical force element is

d\tau_z = x\,dF_y - y\,dF_x = 0

if force were only along $z$ . So instead it is easier to use the magnetic dipole approximation since $b\gg a$ .

Dipole moment of the square loop

Area of loop:

A=(2a)^2=4a^2.

Magnetic dipole moment magnitude:

m=IA=4Ia^2.

Since the loop lies in the $xz$ -plane, its normal is along $\pm \hat y$ . Thus

\vec m = 4Ia^2\,\hat y

(in magnitude; sign only affects direction, not magnitude of torque).

Magnetic field at the centre of the loop

Since $b\gg a$ , field is approximately uniform over the loop and equal to field at the origin.

Distance of origin from the wire is $b$ , so

B=\frac{\mu_0 I}{2\pi b}.

Direction at origin is along $\hat x$ (or $-\hat x$ depending on current direction), perpendicular to $\vec m$ . Hence angle between $\vec m$ and $\vec B$ is $90^\circ$ .

Torque on magnetic dipole

Magnitude of torque is

\tau = mB\sin 90^\circ = mB.

So,

\tau = (4Ia^2)\left(\frac{\mu_0 I}{2\pi b}\right) =\frac{2\mu_0 I^2 a^2}{\pi b}.

Match with options

\boxed{\tau=\frac{2\mu_0 I^2 a^2}{\pi b}}

This corresponds to Option A.

Comparison with stored answer

Stored correct answer: A
Derived answer: A

So they agree.

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