A metallic rod of mass per unit length 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

CONCEPT:

The force on the current-carrying conductor is written as,

F = I \((\vec l \times \vec B) \)

Here, F is the force, I is current, l is the length and B is the magnetic field.

CALCULATION:

The force on the current-carrying conductor is written as;

F = I \((\vec l \times \vec B) \)

Here in the figure, we can see that the magnetic field is in the y-direction. So using Fleming's right-hand rule the force which is applied to the conductor is written as;

\(F = IlBcosθ\) ---(1)

θ = 30 ° (given)

On putting the value of the angle \((θ)\) in equation (1) we have,

F = ILB cos 30°

Also,\(F = mgsin(θ)\) -----(2)

again, on putting the value of the angle \((θ)\) in equation (2) we have,

F = mg sin 30°

For equilibrium, 

mg sin 30° = ILB cos 30° 

\(I = \frac{{mg}}{{lB}}\)tan 30° 

\(= \frac{{0.5 \times 9.8}}{{0.25 \times \sqrt 3 }}\)

= 11.32 A

Hence, option 3) is the correct answer.