To solve the problem of finding the force opposing the motion of an infinitesimal bar magnet with dipole moment M moving with speed v in the x-direction, we need to consider the electromagnetic effects induced in the closed circular loop placed in the x-y-z plane. The loop's resistance is R, and we will assume that the distance x from the magnet to the loop is much greater than R, which simplifies our calculations.
Understanding the Induced EMF
As the bar magnet moves, it creates a changing magnetic field in the vicinity of the loop. According to Faraday's law of electromagnetic induction, a changing magnetic flux through a closed loop induces an electromotive force (EMF) in that loop. The induced EMF (ε) can be expressed as:
ε = -dΦ/dt
where Φ is the magnetic flux through the loop. The negative sign indicates the direction of the induced EMF opposes the change in magnetic flux, as per Lenz's law.
Calculating the Magnetic Flux
The magnetic field (B) due to a dipole moment M at a distance x along the axis of the dipole is given by:
B(x) = (μ₀ / 4π) * (2M / x³)
Here, μ₀ is the permeability of free space. The magnetic flux (Φ) through the circular loop of radius a can be calculated as:
Φ = ∫ B · dA
Since the magnetic field is uniform across the loop at a large distance, we can simplify this to:
Φ = B(x) * A
where A = πa² is the area of the loop. Thus, we have:
Φ = (μ₀ / 4π) * (2M / x³) * πa² = (μ₀Ma²) / (2x³)
Finding the Induced EMF
Now, we differentiate the magnetic flux with respect to time to find the induced EMF:
ε = -dΦ/dt = -d/dt[(μ₀Ma²) / (2x³)]
Since x is changing with time as the magnet moves, we can express x as:
x(t) = vt
Thus, we have:
ε = -d/dt[(μ₀Ma²) / (2(vt)³)] = - (μ₀Ma²) / 2 * d/dt[(vt)⁻³]
Using the chain rule, we find:
d/dt[(vt)⁻³] = -3(vt)⁻⁴ * v = -3v / (v³t³)
Substituting this back, we get:
ε = (3μ₀Ma²v) / (2v³t³) = (3μ₀Ma²) / (2v²t³)
Calculating the Induced Current
The induced current (I) in the loop can be calculated using Ohm's law:
I = ε / R = (3μ₀Ma²) / (2Rv²t³)
Determining the Force Opposing Motion
The force (F) opposing the motion of the magnet can be found using the formula for the magnetic force on a current-carrying loop in a magnetic field:
F = I * B
Substituting the expressions for I and B, we have:
F = (3μ₀Ma²) / (2Rv²t³) * (μ₀ / 4π) * (2M / x³)
Replacing x with vt gives:
F = (3μ₀Ma²) / (2Rv²t³) * (μ₀ / 4π) * (2M / (vt)³)
After simplifying, we find:
F = (3μ₀²M²a²) / (4πRv²t⁶)
Final Expression for the Opposing Force
Thus, the force opposing the motion of the magnet is given by:
F = (3μ₀²M²a²) / (4πRv²t⁶)
This expression shows how the force depends on the dipole moment, the radius of the loop, the resistance, and the speed of the magnet. As time progresses, the force increases due to the t⁶ term in the denominator, indicating that the opposing force becomes more significant as the magnet moves closer to the loop.
