Question

on a smooth plane inclined at 30 degree with the horizontal a thin current carrying matallic rod is placed ll to horizontal field ground the plan is located in uniform magnetic field of 0.157 in the vertical direction for what value of current can the rod remain stationary? the mass per length of rod is 0.03 Kgm-1​

Answer 1

Answer:

The value of current is 1.107 A.

Explanation:

Given data: We have a plane inclined at θ = 30° with the horizontal ground. A current carrying rod with uniform magnetic field, B= 0.157 T, acting in the vertical direction, is place // to ground.

Also, mass per unit length = 0.03 kg m⁻¹

Let “i” be the current flowing through the rod.  

As shown in the figure attached below we can see that the rod across the plane will be affected by a force due to weight i.e., mg in the downward direction. Then a magnetic force, Fm = B* i * L, just // to the horizontal ground, will be acting on the rod. Therefore, the component of the force acting on the rod in the upward direction will be "BiLcos30°" and the component of force acting on the rod in the downward direction will be "mgsin30°".[ taking value of g = 10 m/s^2]

∴ for equilibrium,  

Force upward = Force downward

Or, B*i*L cos30° = mg sin30°

Putting sin30° = ½ and cos30° = √3/2

Or, i = (mg * ½) / (BL * √3/2)

Or, i = 0.03 * 10 / (0.157 * 1 * √3 )

Or, i = 0.3 / 0.271

Or, i = 1.107 A