Problem 5 Solution
1. Identify the Flux:
To find the interaction force, we first calculate the current induced in the ring. The magnetic flux $\Phi$ through the ring due to the magnetic dipole $m$ at distance $x$ on the axis is given by the solid angle formula or by integrating the field. A convenient result for the flux through a circular loop of radius $a$ due to a dipole $m$ on its axis is: $$ \Phi = \frac{\mu_0 m a^2}{2(a^2 + x^2)^{3/2}} $$
To find the interaction force, we first calculate the current induced in the ring. The magnetic flux $\Phi$ through the ring due to the magnetic dipole $m$ at distance $x$ on the axis is given by the solid angle formula or by integrating the field. A convenient result for the flux through a circular loop of radius $a$ due to a dipole $m$ on its axis is: $$ \Phi = \frac{\mu_0 m a^2}{2(a^2 + x^2)^{3/2}} $$
2. Calculate Induced EMF and Current:
As the magnet moves with velocity $v = -\frac{dx}{dt}$ (distance decreasing), the flux changes. The induced EMF is: $$ \mathcal{E} = \left| \frac{d\Phi}{dt} \right| = \left| \frac{d\Phi}{dx} \cdot \frac{dx}{dt} \right| $$ Differentiating $\Phi$ with respect to $x$: $$ \frac{d\Phi}{dx} = \frac{\mu_0 m a^2}{2} \cdot \left( -\frac{3}{2} \right)(a^2 + x^2)^{-5/2} \cdot (2x) = -\frac{3\mu_0 m a^2 x}{2(a^2 + x^2)^{5/2}} $$ Thus, the EMF magnitude is: $$ \mathcal{E} = \frac{3\mu_0 m a^2 x v}{2(a^2 + x^2)^{5/2}} $$ The induced current $I$ in the ring (resistance $R$) is: $$ I = \frac{\mathcal{E}}{R} = \frac{3\mu_0 m a^2 x v}{2R(a^2 + x^2)^{5/2}} $$
As the magnet moves with velocity $v = -\frac{dx}{dt}$ (distance decreasing), the flux changes. The induced EMF is: $$ \mathcal{E} = \left| \frac{d\Phi}{dt} \right| = \left| \frac{d\Phi}{dx} \cdot \frac{dx}{dt} \right| $$ Differentiating $\Phi$ with respect to $x$: $$ \frac{d\Phi}{dx} = \frac{\mu_0 m a^2}{2} \cdot \left( -\frac{3}{2} \right)(a^2 + x^2)^{-5/2} \cdot (2x) = -\frac{3\mu_0 m a^2 x}{2(a^2 + x^2)^{5/2}} $$ Thus, the EMF magnitude is: $$ \mathcal{E} = \frac{3\mu_0 m a^2 x v}{2(a^2 + x^2)^{5/2}} $$ The induced current $I$ in the ring (resistance $R$) is: $$ I = \frac{\mathcal{E}}{R} = \frac{3\mu_0 m a^2 x v}{2R(a^2 + x^2)^{5/2}} $$
3. Calculate the Force:
We can use the conservation of energy principle. The mechanical power done by the external force pushing the magnet (against the repulsive magnetic force $F$) is dissipated as electrical power in the ring. $$ P_{mech} = P_{elec} $$ $$ F \cdot v = I^2 R $$ Substituting the expression for $I$: $$ F = \frac{I^2 R}{v} = \frac{1}{v} \left( \frac{3\mu_0 m a^2 x v}{2R(a^2 + x^2)^{5/2}} \right)^2 R $$ $$ F = \frac{1}{v} \frac{9 \mu_0^2 m^2 a^4 x^2 v^2}{4 R^2 (a^2 + x^2)^5} R $$
We can use the conservation of energy principle. The mechanical power done by the external force pushing the magnet (against the repulsive magnetic force $F$) is dissipated as electrical power in the ring. $$ P_{mech} = P_{elec} $$ $$ F \cdot v = I^2 R $$ Substituting the expression for $I$: $$ F = \frac{I^2 R}{v} = \frac{1}{v} \left( \frac{3\mu_0 m a^2 x v}{2R(a^2 + x^2)^{5/2}} \right)^2 R $$ $$ F = \frac{1}{v} \frac{9 \mu_0^2 m^2 a^4 x^2 v^2}{4 R^2 (a^2 + x^2)^5} R $$
Final Answer:
$$ F = \frac{9 \mu_0^2 a^4 m^2 x^2 v}{4 R (a^2 + x^2)^5} \quad (\text{Repulsive}) $$