EMI BYU 19 - CHATUP707

Solution to Problem 19

Solution: Square Frame in Non-Uniform Magnetic Field

1. Flux and Induced Current

The magnetic field is $\vec{B} = B_0(1+kx) \hat{j}$. The flux through a square loop of side $l$ at position $x$ is: $$ \Phi = \int B dA = B_{avg} l^2 $$ Taking the variation into account properly, the part of the flux that changes with position $x$ is $\Phi(x) \propto B_0 k x \cdot l^2$. The induced EMF due to motion is $\varepsilon = – \frac{d\Phi}{dt} = – B_0 k l^2 v$. Using the inductance equation $L \frac{di}{dt} = \varepsilon$: $$ L \frac{di}{dt} = – B_0 k l^2 \frac{dx}{dt} $$ Integrating: $$ i = – \frac{B_0 k l^2}{L} x $$

2. Force Calculation

The force on a current loop (dipole) in a non-uniform field is $F = \nabla (\vec{\mu} \cdot \vec{B})$ or $F = \mu \frac{dB}{dx}$. Magnetic moment $\mu = i l^2$. Field gradient $\frac{dB}{dx} = B_0 k$. $$ F = (i l^2) (B_0 k) = \left( – \frac{B_0 k l^2}{L} x \right) l^2 B_0 k $$ $$ F = – \frac{B_0^2 k^2 l^4}{L} x $$

3. Motion

This is SHM with effective spring constant $K_{eff} = \frac{B_0^2 k^2 l^4}{L}$. Angular frequency $\omega = \sqrt{\frac{K_{eff}}{m}} = \frac{B_0 k l^2}{\sqrt{mL}}$. The position varies as:

        $$ x(t) = \frac{v_0 \sqrt{mL}}{B_0 k l^2} \sin\left( \frac{B_0 k l^2}{\sqrt{mL}} t \right) $$
    