Question

A metallic rod of mass per unit length $0.5kg/m$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by a current flowing through it when a magnetic field of induction $0.25T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is?
A. $14.76A$
B. $7.14A$
C. $11.32A$
D. $5.98A$

Answer 1

This problem is a question of finding the force on a current carrying conductor. The first most step will be to break down the horizontal components of all the forces in the system of rod and inclined plane. After that we will equate the forces in the two directions of the rod sliding upward or downward, as the rod is not allowed to move.

Formula used:
Force on a current carrying conductor: $F = i(\vec l \times \vec B) = ilB\cos \theta$
Where $F$ is the force on a current carrying rod and is expressed in Newton $(N)$, $B$ is the intensity of magnetic field acting perpendicularly on the rod and is expressed in Webers $(W)$, $l$ is the length of the rod and is expressed in meter $(m)$, $i$ is the current flowing in the rod and is expressed in Amperes $(A)$ and $\theta$ is the angle between the inclination of the plane and the horizontal and is expressed in Radians $(rad)$.
Weight of rod: $mg$
Where $m$ is the mass of the rod and is expressed in kilogram $(kg)$ and $g$ is the acceleration due to gravity and is approximated to be $9.8m/{s^2}$.

Complete step by step answer:
In the above we can see the following components of force acting on the rod:

Vertically:
Magnetic field $B$ (given) perpendicularly upwards.
Weight of rod $mg$ perpendicularly downwards.
Horizontally:
Force on the conductor $F = ilb$ at an angle of $30^\circ$ with the horizontal (Fleming’s Left Hand Rule).
Along the inclination:
Horizontal component of weight of rod $mg\sin \theta$ at an angle of $30^\circ$.
Horizontal component of force on the rod $ilB\cos \theta$ at an angle of $30^\circ$.

The boundary condition given in the question is that the rod does not move and current $i$ is flown through it to ensure this.
Therefore the two opposing sliding forces will be equal. That is,
$mg\sin \theta = ilB\cos \theta$.
Upon rearrangement we get,
$i = \dfrac{{mg}}{{lB}}\tan \theta$.
We know the values of $m/l,g,B,\theta$. Upon substitution we get,
$i = \dfrac{{mg}}{{lB}}\tan \theta = \dfrac{{0.5 \times 9.8}}{{0.25}} \times \dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow i \approx 11.32A \\\$

So, the correct answer is “Option C”.

Note:
Fleming’s left hand rule states that if the thumb denotes the force, index the magnetic field then the middle finger will denote the direction of current. Using this rule we determined the direction of force on the rod.