An infinitesimally small bar magnet of dipole moment M is pointing and moving with the speed v in th

Q: Why do we assume the magnetic field is uniform over the loop?

Because the problem states $x \gg a$, the variation of the magnetic field across the small radius of the loop is negligible, allowing us to treat the field as uniform.

Q: Why is the induced dipole moment parallel to the magnet's dipole moment?

As the magnet moves away, the magnetic flux through the loop decreases. By Lenz's Law, the induced current creates a magnetic field that tries to oppose this decrease, meaning it points in the same direction as the original field.

Q: Can we find the force using Lorentz force instead of potential energy?

Yes! You can calculate the force on the induced current loop using $F = \int I (dl \times B)$. Since the field has a radial component off-axis, this will yield the exact same result, but the potential energy method is mathematically much more elegant.

Q: What does the negative sign in the force derivative mean?

The negative sign indicates that the force is attractive, pointing in the $-x$ direction. This perfectly aligns with Lenz's Law, as the loop tries to pull the receding magnet back to oppose its motion.

Q: Why does the force depend on $x^{-8}$?

The magnetic field of a dipole falls off as $x^{-3}$. The induced current depends on the rate of change of this field, introducing another $x^{-4}$ factor. The energy is their product ($x^{-7}$), and the force is the derivative of energy, leading to the incredibly steep $x^{-8}$ dependence!

Question

An infinitesimally small bar magnet of dipole moment $M$ is pointing and moving with the speed $v$ in the positive $x$-direction. A small closed circular conducting loop of radius $a$ and negligible self-inductance lies in the $y$-$z$ plane with its centre at $x = 0$, and its axis coinciding with the $X$-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is $R$. Assume that the distance $x$ of the magnet from the centre of the loop is much greater than $a$.

Sigma Percentile · Accepted Answer

The problem of a moving magnet and a conducting loop is a classic demonstration of Lenz's Law and Faraday's Law of Induction. But instead of just calculating the induced current, we are asked to find the **mechanical force** opposing the magnet's motion. This requires us to bridge the gap between electromagnetism and mechanics using the concept of magnetic dipoles and potential energy.

### Analyzing the Setup

Imagine a small conducting loop of radius $a$ resting in the $Y$-$Z$ plane. A tiny bar magnet with a dipole moment $M$ is located at a distance $x$ along the $X$-axis. The magnet is moving away from the loop with a constant speed $v$.

Because the magnet is moving, the magnetic field it creates at the location of the loop is constantly changing. This changing magnetic field means the magnetic flux through the loop is also changing. According to Faraday's Law, this will induce an electromotive force (EMF) in the loop.

### The Master Equation

First, we need to determine the magnetic field $B$ produced by the bar magnet at the center of the loop. Since the loop lies on the axial line of the magnet, we use the standard formula for the axial magnetic field of a dipole:

$B = \frac{\mu_0}{4\pi} \frac{2M}{x^3} = \frac{\mu_0 M}{2\pi x^3}$

Because the loop is very small ($x \gg a$), we can safely assume that this magnetic field is uniform over the entire area of the loop. The magnetic flux $\phi$ linked with the loop is simply the product of the magnetic field and the area of the loop ($\pi a^2$):

$\phi = B \cdot A = \left( \frac{\mu_0 M}{2\pi x^3} \right) (\pi a^2) = \frac{\mu_0 M a^2}{2 x^3}$

Now, we apply Faraday's Law to find the induced EMF. We need to differentiate the flux with respect to time. Since the magnet is moving, the distance $x$ is a function of time, and its rate of change $\frac{dx}{dt}$ is exactly the speed $v$.

$e = \left| \frac{d\phi}{dt} \right| = \frac{\mu_0 M a^2}{2} \left| \frac{d}{dt}(x^{-3}) \right|$

Using the chain rule, we get:

$e = \frac{\mu_0 M a^2}{2} \left( \frac{3}{x^4} \right) \frac{dx}{dt} = \frac{3 \mu_0 M a^2 v}{2 x^4}$

### Final Calculation

This induced EMF drives a current $i$ through the loop, which has a resistance $R$.

$i = \frac{e}{R} = \frac{3 \mu_0 M a^2 v}{2 R x^4}$

A current-carrying loop acts as a magnetic dipole itself! The induced dipole moment $M'$ is the product of the induced current and the area of the loop:

$M' = i \cdot (\pi a^2) = \frac{3 \pi \mu_0 M a^4 v}{2 R x^4}$

We now have two interacting magnetic dipoles: the original magnet $M$ and the induced dipole $M'$. By Lenz's Law, the induced current will try to oppose the change in flux. Since the magnet is moving away, the loop will try to attract it. This means the induced dipole moment $M'$ aligns parallel to the magnetic field $B$.

The potential energy $U$ of the induced dipole in the magnetic field is given by:

$U = -M' B = -\left( \frac{3 \pi \mu_0 M a^4 v}{2 R x^4} \right) \left( \frac{\mu_0 M}{2\pi x^3} \right) = -\frac{3 \mu_0^2 M^2 a^4 v}{4 R x^7}$

Finally, the force $F$ between the two dipoles is the negative gradient of their potential energy. We differentiate $U$ with respect to $x$:

$F = \left| -\frac{dU}{dx} \right| = \frac{3 \mu_0^2 M^2 a^4 v}{4 R} \left( \frac{7}{x^8} \right) = \frac{21 \mu_0^2 M^2 a^4 v}{4 R x^8}$

This incredibly steep $x^{-8}$ dependence shows that the magnetic drag force drops off extremely rapidly as the magnet moves away!

Visualized Solution

$Visualizing the Setup$

$Magnetic Field of the Magnet$

$Magnetic Flux through the Loop$

$Faraday’s Law of Induction$

$Evaluating the EMF$

$Induced Current and Dipole Moment$

$Potential Energy of the System$

$Force Opposing the Motion$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A bar magnet is passing through a conducting loop of radius $R$ with velocity $v$ . The radius of the bar magnet is such that it just passes through the loop. The induced emf in the loop can be represented by the approximate curve

List-I

List-II

Comprehension Passage

When the dipole m is placed at a distance r from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to

The work done in bringing the dipole from infinity to a distance r from the center of the loop by the given process is proportional to

Comprehension Passage

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Flux through the Loop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Faraday’s Law of Induction

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Induced Current and Dipole Moment

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

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

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

List-I

List-II

Comprehension Passage

When the dipole m is placed at a distance r from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to

The work done in bringing the dipole from infinity to a distance r from the center of the loop by the given process is proportional to

Comprehension Passage

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

$Visualizing the Setup$

$Magnetic Field of the Magnet$

$Magnetic Flux through the Loop$

$Faraday’s Law of Induction$

$Evaluating the EMF$

$Induced Current and Dipole Moment$

$Potential Energy of the System$

$Force Opposing the Motion$