Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each oth

Q: Why is the magnetic field of Y upwards?

The dipole moment of Y points downwards (from South to North). Since point P lies on the equatorial line of Y, the magnetic field is anti-parallel to the dipole moment, hence it points upwards.

Q: Why do we use the short dipole approximation?

The problem states that 'd is much larger than the dimensions of the dipole', which is the exact condition for using the short dipole formulas $2M/r^3$ and $M/r^3$.

Q: What if the velocity was along the x-axis?

If the velocity was along the x-axis, the angle between $\vec{v}$ and $\vec{B}_{net}$ would be $45^\circ$, and the force would be $qvB_{net}\sin(45^\circ)$.

Q: Why is the force zero when velocity is parallel to the magnetic field?

The magnetic force is given by the cross product $\vec{v} \times \vec{B}$. The magnitude of a cross product is $vB\sin(\theta)$. When vectors are parallel, the angle $\theta$ is $0^\circ$, and $\sin(0^\circ) = 0$.

Q: Does the charge of the particle affect the direction of the force?

Yes, if the force was non-zero, a negative charge would experience a force in the opposite direction to that given by the right-hand rule. However, since the cross product is zero here, the charge sign doesn't matter.

Question

Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their mid-point P, at angle $	heta = 45^\circ$ with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? (d is much larger than the dimensions of the dipole)

Options:

(A) $\left(\frac{\mu_0}{4\pi}ight) \frac{M}{(d/2)^3} 	imes qv$

(B) 0

(C) $\sqrt{2}\left(\frac{\mu_0}{4\pi}ight) \frac{M}{(d/2)^3} 	imes qv$

(D) $\left(\frac{\mu_0}{4\pi}ight) \frac{2M}{(d/2)^3} 	imes qv$

Sigma Percentile · Accepted Answer

The problem of finding the magnetic force on a moving charge in the presence of multiple magnetic dipoles might seem daunting at first glance. However, as we will see, a careful analysis of the magnetic fields reveals a beautifully elegant solution. Let's break down the physics step by step.

### Analyzing the Setup

Imagine two magnetic dipoles, **X** and **Y**, placed at a separation of $d$. A charged particle $q$ is passing through their midpoint $P$ with a velocity $\vec{v}$, at an angle of $45^\circ$ to the horizontal.

Our ultimate goal is to find the magnetic force acting on this particle at this exact instant. To do this, we must rely on the Lorentz force formula, which states that the magnetic force on a moving charge is given by:

$$ \vec{F} = q(\vec{v} 	imes \vec{B}_{net}) $$

This equation tells us that the force depends on the charge, its velocity, and the net magnetic field at point $P$. Therefore, our very first task is to determine the magnitude and direction of this net magnetic field, $\vec{B}_{net}$.

### The Magnetic Field of Dipole X

Let's focus on dipole **X** first. Point $P$ lies exactly on its axial line. We know from the theory of magnetism that the magnetic field on an axial line is directed along the dipole moment, which always points from the South pole to the North pole.

Since dipole **X** is oriented horizontally with its North pole facing right, its dipole moment $\vec{M}_X$ points to the right. Consequently, the magnetic field $\vec{B}_X$ at point $P$ also points to the right.

Using the formula for the magnetic field of a short dipole on its axis, its magnitude is:

$$ B_X = \frac{\mu_0}{4\pi} \frac{2M}{(d/2)^3} $$

### The Magnetic Field of Dipole Y

Next, let's look at dipole **Y**. Point $P$ lies on its equatorial line. At an equatorial position, the magnetic field is always directed opposite to the dipole moment.

Dipole **Y** is oriented vertically with its South pole up and North pole down, meaning its dipole moment $\vec{M}_Y$ points downwards. Therefore, the magnetic field $\vec{B}_Y$ at point $P$ must point upwards.

Now, notice a crucial detail: the dipole moment of **Y** is $2M$. According to the equatorial formula, the magnitude of the field is:

$$ B_Y = \frac{\mu_0}{4\pi} \frac{2M}{(d/2)^3} $$

Fascinatingly, this magnitude is exactly equal to $B_X$!

### The Net Magnetic Field

So, we now have two magnetic fields at point $P$: $\vec{B}_X$ pointing horizontally to the right, and $\vec{B}_Y$ pointing vertically upwards. Furthermore, we established that their magnitudes are perfectly equal ($|\vec{B}_X| = |\vec{B}_Y|$).

When two vectors are equal in magnitude and perpendicular to each other, their resultant perfectly bisects the angle between them. This means our net magnetic field, $\vec{B}_{net}$, will point at exactly $45^\circ$ above the horizontal.

### The Lorentz Force and the Final Catch

Here is where the beautiful catch of the problem lies! The question states that the particle's velocity $\vec{v}$ is also directed at $45^\circ$ above the horizontal.

This means the velocity vector $\vec{v}$ and the net magnetic field vector $\vec{B}_{net}$ are perfectly parallel to each other.

$$ \vec{v} \parallel \vec{B}_{net} $$

As we know from vector algebra, the cross product of two parallel vectors is always zero.

$$ \vec{v} 	imes \vec{B}_{net} = 0 $$

Therefore, substituting this back into our Lorentz force equation, the magnetic force acting on the particle is simply zero.

$$ \vec{F} = 0 $$

It's an elegant result that rewards a solid conceptual understanding over brute-force calculation!

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Visualized Solution

$System Setup$

$Lorentz Force$

$Magnetic Field due to X$

$Magnetic Field due to Y$

$Net Magnetic Field$

$Calculating Force$

The Sigma Insight: Bar Magnet

Analyzing the Setup

The Magnetic Field of Dipole X

The Magnetic Field of Dipole Y

The Net Magnetic Field

The Lorentz Force and the Final Catch

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lorentz Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field due to X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field due to Y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Force