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

To determine the current required to keep the metallic rod stationary on the inclined plane, we apply equilibrium of forces and magnetic force.

1. The rod is on an inclined plane making an angle of \( 30^\circ \) with the horizontal. A component of gravitational force acts along the plane.
2. Force due to gravity along the incline:

   \[ F_{\text{gravity}} = m g \sin\theta \]

   where:
   • \( m = 0.5\,\text{kg/m} \)
   • \( g = 9.8\,\text{m/s}^2 \)
   • \( \theta = 30^\circ \)

   Substituting values:

   \[ F_{\text{gravity}} = 0.5 \times 9.8 \times \sin 30^\circ = 0.5 \times 9.8 \times 0.5 = 2.45\,\text{N/m} \]

3. Magnetic force on the rod:

   \[ F_B = I L B \sin\phi \]

   where:
   • \( I \) = current
   • \( L = 1 \) (per unit length)
   • \( B = 0.25\,\text{T} \)
   • \( \phi = 90^\circ \)

   Since \( \sin 90^\circ = 1 \):

   \[ F_B = I \times 0.25 \]

4. For equilibrium:

   \[ F_B = F_{\text{gravity}} \]

   \[ I \times 0.25 = 2.45 \]

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

5. Thus, the current required is \( 9.8\,\text{A} \).