A wire loop carrying a current I is placed in the x-y plane as shown in figure.
(a) If a particle wi

Q: Why is the magnetic field due to the chord pointing inwards?

Using the right-hand grip rule, if you point your thumb in the direction of the current along the chord (from N to M, which is upwards), your fingers curl inwards into the page at the origin P.

Q: How did we determine the direction of the velocity vector?

The problem states the velocity is along NP. Point N is at 240 degrees on the circle. The vector from N to the center P points in the opposite direction, which is 240 - 180 = 60 degrees relative to the positive x-axis.

Q: Why is the net force on the loop zero in part (b)?

In a uniform magnetic field, the force on any current element $d\mathbf{l}$ is $I (d\mathbf{l} \times \mathbf{B})$. When you integrate this over a closed loop, the integral of $d\mathbf{l}$ is zero, so the net force is always zero, regardless of the loop's shape.

Q: How do we calculate the area of this specific loop shape?

The loop forms a circular segment. Its area is the area of the circular sector (the "pie slice" from P to the arc) minus the area of the triangle formed by P, M, and N.

Q: Why does the torque point in the y-direction?

The magnetic moment $\mathbf{M}$ is in the z-direction ($\hat{\mathbf{k}}$) because the current is counter-clockwise. The external field $\mathbf{B}$ is in the x-direction ($\hat{\mathbf{i}}$). The cross product $\hat{\mathbf{k}} \times \hat{\mathbf{i}}$ gives $\hat{\mathbf{j}}$, which is the y-direction.

Question

A wire loop carrying a current $I$ is placed in the $x$-$y$ plane as shown in figure.
(a) If a particle with charge $+Q$ and mass $m$ is placed at the centre $P$ and given a velocity $v$ along $NP$ (see figure), find its instantaneous acceleration.
(b) If an external uniform magnetic induction field $\mathbf{B} = B \hat{\mathbf{i}}$ is applied, find the force and the torque acting on the loop due to this field.

Sigma Percentile · Accepted Answer

This problem is a beautiful synthesis of multiple core concepts in electromagnetism: the Biot-Savart law, the Lorentz force, and the behavior of magnetic dipoles in external fields. Let's break it down step-by-step.

### Analyzing the Setup

We are given a current loop in the $x$-$y$ plane. The loop is composed of two distinct geometric segments: a circular arc $MN$ and a straight chord $NM$. The center of the circle, $P$, is placed at the origin $(0,0)$. The arc subtends an angle of $120^\circ$ at the center.

By observing the current arrows, we see that the current $I$ flows counter-clockwise along the minor arc from $M$ to $N$, and then straight up along the chord from $N$ back to $M$. This counter-clockwise flow is crucial for determining the direction of the magnetic fields and the magnetic dipole moment.

### The Master Equation: Net Magnetic Field

To find the instantaneous acceleration of a charged particle placed at the center, we first need to determine the net magnetic field at $P$. We use the principle of superposition, calculating the field from the arc and the chord separately.

**1. Field from the Circular Arc:**
The magnetic field at the center of a full circular loop is $\frac{\mu_0 I}{2a}$. Since our arc subtends $120^\circ$, which is exactly $\frac{1}{3}$ of a full circle, its contribution is proportionally scaled:

$$\mathbf{B}_1 = \frac{1}{3} \left( \frac{\mu_0 I}{2a} ight) \hat{\mathbf{k}} = \frac{\mu_0 I}{6a} \hat{\mathbf{k}} \approx 0.16 \frac{\mu_0 I}{a} \hat{\mathbf{k}}$$

Using the right-hand grip rule, the counter-clockwise current produces a field pointing outwards, in the positive $z$-direction ($\hat{\mathbf{k}}$).

**2. Field from the Straight Chord:**
The chord is a finite straight wire. The perpendicular distance from the center $P$ to the chord is $r = a \cos 60^\circ = \frac{a}{2}$. The ends of the chord subtend angles $	heta_1 = 60^\circ$ and $	heta_2 = 60^\circ$ at $P$. Applying the Biot-Savart law for a finite wire:

$$\mathbf{B}_2 = \frac{\mu_0 I}{4\pi r} (\sin 	heta_1 + \sin 	heta_2) (-\hat{\mathbf{k}})$$

$$\mathbf{B}_2 = \frac{\mu_0 I}{4\pi (a/2)} (2 \sin 60^\circ) (-\hat{\mathbf{k}}) = -\frac{\sqrt{3} \mu_0 I}{2\pi a} \hat{\mathbf{k}} \approx -0.27 \frac{\mu_0 I}{a} \hat{\mathbf{k}}$$

The upward current in the chord produces a field pointing inwards, in the negative $z$-direction.

**3. Net Magnetic Field:**
Adding these contributions vectorially gives the net magnetic field at the center:

$$\mathbf{B}_{	ext{net}} = \mathbf{B}_1 + \mathbf{B}_2 = \left( \frac{1}{6} - \frac{\sqrt{3}}{2\pi} ight) \frac{\mu_0 I}{a} \hat{\mathbf{k}} \approx -0.11 \frac{\mu_0 I}{a} \hat{\mathbf{k}}$$

### Lorentz Force and Acceleration

A particle with charge $+Q$ is projected from $P$ along the line $NP$. Since $N$ is at an angle of $240^\circ$, the direction from $N$ to $P$ is $240^\circ - 180^\circ = 60^\circ$ relative to the positive $x$-axis. The velocity vector is:

$$\mathbf{v} = v \cos 60^\circ \hat{\mathbf{i}} + v \sin 60^\circ \hat{\mathbf{j}} = \frac{v}{2} \hat{\mathbf{i}} + \frac{\sqrt{3}v}{2} \hat{\mathbf{j}}$$

The magnetic force is given by the Lorentz force law, $\mathbf{F}_m = Q (\mathbf{v} 	imes \mathbf{B}_{	ext{net}})$. Taking the cross product:

$$\mathbf{F}_m = Q \left( \frac{v}{2} \hat{\mathbf{i}} + \frac{\sqrt{3}v}{2} \hat{\mathbf{j}} ight) 	imes \left( -0.11 \frac{\mu_0 I}{a} \hat{\mathbf{k}} ight)$$

$$\mathbf{F}_m = \frac{0.11 \mu_0 I Q v}{2a} \hat{\mathbf{j}} - \frac{0.11 \sqrt{3} \mu_0 I Q v}{2a} \hat{\mathbf{i}}$$

Dividing by the mass $m$ yields the instantaneous acceleration:

**$$\mathbf{a} = \frac{0.11 \mu_0 I Q v}{2am} (\hat{\mathbf{j}} - \sqrt{3} \hat{\mathbf{i}})$$**

### The Loop in an External Field

In the second part of the problem, the entire loop is placed in a uniform external magnetic field $\mathbf{B} = B \hat{\mathbf{i}}$.

A fundamental law of electromagnetism states that the net magnetic force on *any* closed current loop in a uniform magnetic field is exactly zero. Thus, **$\mathbf{F} = 0$**.

However, the loop will experience a torque, $\boldsymbol{	au} = \mathbf{M} 	imes \mathbf{B}$. To find this, we must calculate the magnetic dipole moment $\mathbf{M} = I A \hat{\mathbf{k}}$. The area $A$ of our loop is the area of the circular sector minus the area of the triangle $PMN$:

$$A = \frac{1}{3} (\pi a^2) - \frac{1}{2} (2a \sin 60^\circ)(a \cos 60^\circ) = a^2 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} ight) \approx 0.61 a^2$$

So, the magnetic moment is $\mathbf{M} = 0.61 I a^2 \hat{\mathbf{k}}$. Finally, we compute the torque:

$$\boldsymbol{	au} = (0.61 I a^2 \hat{\mathbf{k}}) 	imes (B \hat{\mathbf{i}}) = 0.61 I a^2 B (\hat{\mathbf{k}} 	imes \hat{\mathbf{i}})$$

**$$\boldsymbol{	au} = 0.61 I a^2 B \hat{\mathbf{j}}$$**

This torque will tend to rotate the loop around the $y$-axis, aligning its magnetic moment with the external field.

Visualized Solution

Geometry of the Current Loop

Magnetic Field due to the Arc

Magnetic Field due to the Chord

Net Magnetic Field at the Center

Velocity Vector of the Particle

Magnetic Force and Acceleration

Loop in an External Magnetic Field

Magnetic Dipole Moment of the Loop

Torque on the Loop

The Sigma Insight: Magnetic Force on Current

Analyzing the Setup

The Master Equation

Lorentz Force and Acceleration

The Loop in an External Field

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