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 the z-axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about zaxis is given by :

Magnetics PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q55 with Solution

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Correct answer: A

Geometry of the setup

The square loop lies in the $XZ$ -plane, so its area vector is along $\pm \hat y$ .
Side of square $=2a$ , so area $A=(2a)^2=4a^2.$
The long straight wire is parallel to the $z$ -axis and passes through $(0,b,0)$ .
Since $b\gg a$ , the magnetic field due to the wire is nearly uniform over the loop.

Magnetic field at the centre of the loop

Distance of the origin from the wire is $b$ .

Magnetic field due to a long straight wire: $B=\frac{\mu_0 I}{2\pi b}.$

Now determine its direction at the origin.

The wire is along $z$ and located at $y=b$ . At the origin, the position relative to the wire is toward $-\hat y$ . By right-hand rule, the field there is along $+\hat x$ (or $-\hat x$ depending on current direction, but for magnitude only this does not matter).

So the field is perpendicular to the magnetic moment of the loop.

Magnetic moment of the loop

Magnitude of magnetic moment: $m=IA=I(4a^2)=4Ia^2.$

Direction is normal to the loop, i.e. along $\pm \hat y$ .

Torque on a current loop

For a current loop in a uniform magnetic field, $\tau = mB\sin\theta,$ where $\theta$ is the angle between $\vec m$ and $\vec B$ .

Here $\vec m \parallel \hat y$ and $\vec B \parallel \hat x$ , so $\theta=90^\circ, \qquad \sin\theta=1.$

Hence

=\frac{2\mu_0 I^2 a^2}{\pi b}.$$ --- 5. **Torque about the $z$-axis** Since $\vec \tau = \vec m \times \vec B$, with $\vec m$ along $\hat y$ and $\vec B$ along $\hat x$, the torque is along the $z$-axis. Therefore the magnitude of torque about the $z$-axis is the same as above: $$\boxed{\tau_z=\frac{2\mu_0 I^2 a^2}{\pi b}}.$$ --- 6. **Option check** - **A:** $\dfrac{2\mu_0 I^2 a^2}{\pi b}$ ✅ - **B:** $\dfrac{\mu_0 I^2 a^2}{2\pi b}$ ❌ - **C:** $\dfrac{\mu_0 I^2 a^3}{2\pi b^2}$ ❌ - **D:** $\dfrac{2\mu_0 I^2 a^3}{\pi b^2}$ ❌ So the correct option is **A**.

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View written solutionFree

Correct answer: A

Geometry of the setup

The square loop lies in the $XZ$ -plane, so its area vector is along $\pm \hat y$ .
Side of square $=2a$ , so area $A=(2a)^2=4a^2.$
The long straight wire is parallel to the $z$ -axis and passes through $(0,b,0)$ .
Since $b\gg a$ , the magnetic field due to the wire is nearly uniform over the loop.

Magnetic field at the centre of the loop

Distance of the origin from the wire is $b$ .

Magnetic field due to a long straight wire: $B=\frac{\mu_0 I}{2\pi b}.$

Now determine its direction at the origin.

So the field is perpendicular to the magnetic moment of the loop.

Magnetic moment of the loop

Magnitude of magnetic moment: $m=IA=I(4a^2)=4Ia^2.$

Direction is normal to the loop, i.e. along $\pm \hat y$ .

Torque on a current loop

For a current loop in a uniform magnetic field, $\tau = mB\sin\theta,$ where $\theta$ is the angle between $\vec m$ and $\vec B$ .

Here $\vec m \parallel \hat y$ and $\vec B \parallel \hat x$ , so $\theta=90^\circ, \qquad \sin\theta=1.$

Hence

=\frac{2\mu_0 I^2 a^2}{\pi b}.$$ --- 5. **Torque about the $z$-axis** Since $\vec \tau = \vec m \times \vec B$, with $\vec m$ along $\hat y$ and $\vec B$ along $\hat x$, the torque is along the $z$-axis. Therefore the magnitude of torque about the $z$-axis is the same as above: $$\boxed{\tau_z=\frac{2\mu_0 I^2 a^2}{\pi b}}.$$ --- 6. **Option check** - **A:** $\dfrac{2\mu_0 I^2 a^2}{\pi b}$ ✅ - **B:** $\dfrac{\mu_0 I^2 a^2}{2\pi b}$ ❌ - **C:** $\dfrac{\mu_0 I^2 a^3}{2\pi b^2}$ ❌ - **D:** $\dfrac{2\mu_0 I^2 a^3}{\pi b^2}$ ❌ So the correct option is **A**.

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