Q: A magnetic field $ B = \frac{B_0 y}{a} \hat{k} $ is directed into the paper in the + Z direction. B_{0} and ‘ a ‘ are the positive constants . A square loop EFGH of side ‘ a ‘ mass m and resistance R , in x-y plane, starts falling under the influence of gravity Find

(a) the induced current in the loop and indicate its direction

(b) the total Lorentz force acting on the loop and indicate its direction

(c) an expression for the speed of the loop v (t) and its terminal value.

Solution: (a) Let v = velocity of the loop at any time t .

The induced emf between E & F

$\displaystyle e_1 = B_1 l v = \frac{B_0 y}{a} l v $

$\displaystyle e_1 = B_0 y v $

The induced emf between H and G

$\displaystyle e_2 = B_2 l v = \frac{B_0 (y+a)}{a} l v $

$\displaystyle e_1 = B_0 ( y + a ) v $

The net emf e = e_{2} – e_{1} = B_{0} a v

The induced current $\displaystyle I = – \frac{e}{R} = \frac{B_0 a v}{R}$ anticlockwise because e_{2} > e_{1}

(b) The Lorentz force on EF & HG are given as

$\displaystyle F_1 = \frac{B_0 y}{a} I a = B_0 y I$

$\displaystyle F_2 = \frac{B_0 (y+a)}{a} I a = B_0 (y+a) I$

$\displaystyle F_{net} = F_2 – F_1 = B_0 a I$ along negative Y axis

$\displaystyle F_{net} = – B_0 a I \hat{y} $

$\displaystyle F_{net} = – \frac{B_0^2 a^2 v}{R} \hat{y}$

(c) $\displaystyle F_{net} = m g – \frac{B_0^2 a^2 v}{R} \hat{y}$

$\displaystyle m \frac{dv}{dt} = m g – \frac{B_0^2 a^2 v}{R} $

$\displaystyle \int_{0}^{v} \frac{dv}{mg -\frac{B_0^2 a^2 v}{R} } = \int_{0}^{t} \frac{1}{m} dt$

Evaluating the integration and rearranging the terms we get,

$\displaystyle v = \frac{m g R}{B_0^2 a^2} [1 – e^{\frac{B_0^2 a^2 t}{m R}}] $

When t → ∞ , v → v_{terminal}

$\displaystyle v = \frac{m g R}{B_0^2 a^2} $