Q: Two infinite parallel wires, having the cross sectional area a and resistivity k are connected at a junction point P (as shown in the figure). A slide wire of negligible resistance and having mass ‘ m ‘ and length l can slide between the parallel wires, without any frictional resistance.
If the system of wires is introduced to a magnetic field of intensity B (into the plane of paper) and the slide wire is pulled with a force which varies with the velocity of the slide wire as F = F0 V , then find the velocity of the slide wire as a function of the distance travelled. (The slide wire is initially at origin and has a velocity v0 )
Solution : At any given instance of time the slide wire is at distance x from origin, then the resistance of the circuit is $R = \frac{k(2x+l)}{a}$
If the velocity of slide wire is v , then the emf generated is B l v so we have
$\displaystyle B l v – \frac{k(2x+l)}{a} l = 0 $
$\displaystyle I = \frac{B l a}{k}(\frac{v}{2x+l}) $
This current exerts magnetic force in the wire given by F’
$\displaystyle F’ = I l B = \frac{B l a}{k}(\frac{v}{2x+l})l B $
$\displaystyle F’ = \frac{B^2 l^2 a}{k}(\frac{v}{2x+l}) $
Since , $\displaystyle F – F’ = m v \frac{dv}{dx} $
$\displaystyle F_0 – \frac{B^2 l^2 a}{k}(\frac{v}{2x+l}) = m v \frac{dv}{dx} $
$\displaystyle \frac{F_0}{m} – \frac{B^2 l^2 a}{k m}(\frac{1}{2x+l}) = \frac{dv}{dx} $
$\displaystyle \int_{0}^{x} [ \frac{F_0}{m} – \frac{B^2 l^2 a}{k m}(\frac{1}{2x+l}) ] dx = \int_{v_0}^{v} dv $
$\displaystyle \frac{F_0}{m} x – \frac{B^2 l^2 a}{2 k m} [ln (\frac{2x+l}{l})] = v – v_0 $
$\displaystyle v = v_0 + \frac{F_0}{m} x – \frac{B^2 l^2 a}{2 k m} [ln (\frac{2x+l}{l})] $