A uniform constant magnetic field B is directed at an angle of 45 degrees to the x-axis in x-y plane

Q: Why does the loop rotate about the diagonal QS?

Because the net force on the loop in a uniform magnetic field is zero, it rotates about an axis passing through its center of mass. The torque vector determines this axis, which aligns perfectly with the diagonal QS.

Q: How did we find the moment of inertia about the diagonal?

We used the perpendicular axis theorem ($I_z = I_x + I_y$). Since the diagonals are perpendicular and the square is symmetric, $I_{QS} = I_{PR}$, so $2I_{QS} = I_{ZZ}$. We then halved the given z-axis moment of inertia.

Q: Why is the magnetic moment in the positive z-direction?

The current flows anti-clockwise (from P to Q to R to S). Using the right-hand grip rule, curling your fingers in the direction of the current makes your thumb point outwards, which corresponds to the positive z-direction.

Q: Can we use $\tau = MB \sin\theta$ directly to find the torque?

Yes, you can find the magnitude directly since the angle between $\mathbf{M}$ (along z) and $\mathbf{B}$ (in xy-plane) is $90^\circ$. So $|\mathbf{\tau}| = MB \sin 90^\circ = I_0 L^2 B$. However, calculating the vector form is crucial to find the exact axis of rotation.

Q: Why do we use $\Delta\theta = \frac{1}{2} \alpha (\Delta t)^2$?

Because the time interval $\Delta t$ is very short, we can assume the angular acceleration $\alpha$ remains constant during this interval. The initial angular velocity is zero, so the standard kinematic equation simplifies to this form.

Question

A uniform constant magnetic field $\mathbf{B}$ is directed at an angle of $45^\circ$ to the $x$-axis in $x$-$y$ plane. $PQRS$ is rigid square wire frame carrying a steady current $I_0$, with its centre at the origin $O$. At time $t = 0$, the frame is at rest in the position shown in the figure with its sides parallel to $x$ and $y$-axes. Each side of the frame is of mass $M$ and length $L$. (1998)
(a) What is the magnitude of torque $\mathbf{	au}$ acting on the frame due to the magnetic field ?
(b) Find the angle by which the frame rotates under the action of this torque in a short interval of time $\Delta t$ and the axis about which this rotation occurs ($\Delta t$ is so short that any variation in the torque during this interval may be neglected). Given : the moment of inertia of the frame about an axis through its centre perpendicular to its plane is $\frac{4}{3}ML^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's visualize the setup. We have a rigid square wire frame $PQRS$ carrying a steady anti-clockwise current $I_0$. By applying the right-hand grip rule, we can determine that its magnetic moment $\mathbf{M}$ points outwards, directly along the positive z-axis.

The magnitude of the magnetic moment is the product of the current and the area of the loop:

$$\mathbf{M} = I_0 A \hat{\mathbf{k}} = I_0 L^2 \hat{\mathbf{k}}$$

The uniform magnetic field $\mathbf{B}$ is directed at a $45^\circ$ angle in the xy-plane. We can express this field in vector notation as:

$$\mathbf{B} = B \cos 45^\circ \hat{\mathbf{i}} + B \sin 45^\circ \hat{\mathbf{j}} = \frac{B}{\sqrt{2}} (\hat{\mathbf{i}} + \hat{\mathbf{j}})$$

### Calculating the Torque

Now, the torque $\mathbf{	au}$ acting on a magnetic dipole in a uniform magnetic field is given by the cross product of the magnetic moment and the magnetic field. Let's substitute our vectors into this relationship:

$$\mathbf{	au} = \mathbf{M} 	imes \mathbf{B}$$

$$\mathbf{	au} = (I_0 L^2 \hat{\mathbf{k}}) 	imes \left[ \frac{B}{\sqrt{2}} (\hat{\mathbf{i}} + \hat{\mathbf{j}}) ight]$$

Evaluating the cross products, we know that $\hat{\mathbf{k}} 	imes \hat{\mathbf{i}} = \hat{\mathbf{j}}$ and $\hat{\mathbf{k}} 	imes \hat{\mathbf{j}} = -\hat{\mathbf{i}}$. This gives us the torque vector:

$$\mathbf{	au} = \frac{I_0 L^2 B}{\sqrt{2}} (\hat{\mathbf{j}} - \hat{\mathbf{i}})$$

The magnitude of this torque simplifies beautifully:

$$|\mathbf{	au}| = \frac{I_0 L^2 B}{\sqrt{2}} \sqrt{(-1)^2 + 1^2} = I_0 L^2 B$$

### Identifying the Axis of Rotation

Look closely at the direction of the torque vector: $\hat{\mathbf{j}} - \hat{\mathbf{i}}$. This vector points exactly along the diagonal connecting vertices $Q$ and $S$.

Since the net force on a closed current loop in a uniform magnetic field is always zero, the loop will naturally tend to rotate about an axis passing through its center of mass. In this specific case, the axis of rotation aligns perfectly with the diagonal $QS$.

### Moment of Inertia via Symmetry

To find the angular acceleration, we need the moment of inertia of the square frame about this diagonal axis $QS$. We can cleverly use the perpendicular axis theorem. The sum of the moments of inertia about the two diagonals ($QS$ and $PR$) equals the moment of inertia about the z-axis ($ZZ$) passing through the center.

$$I_{QS} + I_{PR} = I_{ZZ}$$

By the inherent symmetry of the square, the moments of inertia about both diagonals must be equal ($I_{QS} = I_{PR}$). Therefore:

$$2 I_{QS} = I_{ZZ}$$

We are given that the moment of inertia about the z-axis is $I_{ZZ} = \frac{4}{3} M L^2$. Substituting this in, we find:

$$I_{QS} = \frac{2}{3} M L^2$$

### Final Kinematics

Finally, let's calculate the angular acceleration $\alpha$ by dividing the torque magnitude by the moment of inertia about the rotation axis:

$$\alpha = \frac{|\mathbf{	au}|}{I_{QS}} = \frac{I_0 L^2 B}{\frac{2}{3} M L^2} = \frac{3 I_0 B}{2 M}$$

Notice how the $L^2$ terms cancel out nicely. For a very short time interval $\Delta t$, we can assume the angular acceleration remains constant. Since the frame starts from rest, the angular displacement $\Delta	heta$ is simply:

$$\Delta	heta = \frac{1}{2} \alpha (\Delta t)^2$$

Substituting our expression for $\alpha$, we arrive at our final elegant answer for the angle of rotation:

$$\Delta	heta = \frac{3 I_0 B}{4 M} (\Delta t)^2$$

Visualized Solution

$\text{Visual Anchor & Setup}$

$Torque Calculation$

$Axis of Rotation$

$Moment of Inertia$

$\text{Angular Acceleration & Kinematics}$

The Sigma Insight: Magnetic Moment of Current Loop

Analyzing the Setup

Calculating the Torque

Identifying the Axis of Rotation

Moment of Inertia via Symmetry

Final Kinematics

Similar Questions

A wire carrying current $I$ is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$ , then the magnitude and direction of magnetic moment of the loop ABCDEFA is

A circular coil having $N$ turns and radius $r$ carries a current $I$ . It is held in the XZ-plane in a magnetic field $B \hat{i}$ . The torque on the coil due to the magnetic field (in N-m) is

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is

An insulating thin rod of length $l$ has a linear charge density $ρ (x) = ρ_{0} \frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin ( $x = 0$ ) and perpendicular to the rod. If the rod makes $n$ rotations per second, then the time averaged magnetic moment of the rod is

A thin circular disk of radius $R$ is uniformly charged with density $σ > 0$ per unit area. The disk rotates about its axis with a uniform angular speed $ω$ . The magnetic moment of the disk is