Question

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

• A. 7.14 A
• B. 14.76 A
• C. 5.98 A
• D. 11.32 A

Answer 1

The Correct Option is D

We need to determine the current required to keep a metallic rod stationary on a smooth inclined plane at an angle of \( 30^\circ \) in a vertical magnetic field.

Step 1: Forces Acting on the Rod

• Gravitational force acting downward.
• Magnetic force due to current in the rod.

Step 2: Component of Gravitational Force

The component of weight along the plane is:

\[ F_g = mg \sin\theta \]

Given mass per unit length \( = 0.5\,\text{kg/m} \), for unit length:

\[ F_g = 0.5 \times 9.8 \times \sin 30^\circ = 0.5 \times 9.8 \times 0.5 = 2.45\,\text{N} \]

Step 3: Magnetic Force

\[ F_B = B I L \]

For unit length \( L = 1 \):

\[ F_B = 0.25 \times I \]

Step 4: Equilibrium Condition

\[ F_B = F_g \]

\[ 0.25 I = 2.45 \]

\[ I = \frac{2.45}{0.25} = 9.8\,\text{A} \]

Conclusion: The required current is \( 9.8\,\text{A} \).