Question

On a smooth plane inclined at ${30}^0$ with the horizontal, a thin current-carrying metallic rod is placed parallel to the horizontal ground. The plane is located in a uniform magnetic field of 0.15 T in the vertical direction. For what value of current can the rod remain stationary? The mass per unit length of the rod is 0.03 kg $m^{-1}$.

Answer 1

The current carrying rod will be stationary when net vector summation all forces acting on body is zero.

Hence the force due to magnetic field acting on current carrying rod is Bil in horizontal direction.

F upward$=B_{l}l\cos 30=B_{i}l\sqrt{3}/2$

F downward the gravity $mg\sin 30=\frac{mg}{2}$

For equIIlibrium

F upward=F downward

$i=\frac{mg}{Bl}\times \sqrt{3}$

$i-=0.03\times \frac{10}{0.15}\times 1\times 1.732$

$i=1.15$Amp.