Metallic Rod Equilibrium on an Inclined Plane - Detailed Solution and Interactive Quiz

Metallic Rod Equilibrium Problem

Question: A metallic rod of mass per unit length \( \lambda = 0.5\,\text{kg/m} \) is lying horizontally on a smooth inclined plane that makes an angle \( 30^\circ \) with the horizontal. The rod is prevented from sliding down by flowing a current through it in the presence of a vertical magnetic field of induction \( B = 0.25\,\text{T} \). Find the current flowing in the rod to keep it stationary.

(a) \( 7.14\,\text{A} \)

(b) \( 5.98\,\text{A} \)

(c) \( 14.76\,\text{A} \)

(d) \( 11.32\,\text{A} \)

Detailed Step-by-Step Explanation

Step 1: Determine the Gravitational Force Component
The rod has a mass per unit length \( \lambda = 0.5\,\text{kg/m} \). Thus, its weight per unit length is:

\( w = \lambda g = 0.5 \times 10 = 5\,\text{N/m} \)

Since the plane is inclined at \( 30^\circ \), the component of the weight acting down the plane is:

\( w_{\parallel} = w \sin30^\circ = 5 \times 0.5 = 2.5\,\text{N/m} \)

Step 2: Magnetic Force on the Rod
The magnetic force on a current-carrying rod of length \( L \) in a magnetic field \( B \) is given by:

\( F_{\text{mag}} = I L B \sin\theta \)

Here, the rod is horizontal and the magnetic field is vertical, so they are perpendicular, and \( \sin90^\circ = 1 \). Thus, the total magnetic force per unit length is:

\( f_{\text{mag}} = I B \)

However, to balance the gravitational component along the inclined plane, only the component of the magnetic force along the plane is effective. If the rod is arranged so that the magnetic force makes an angle \( 30^\circ \) with the vertical, then the component along the plane is:

\( f_{\text{mag, parallel}} = I B \cos30^\circ \)

Step 3: Equate Forces for Equilibrium
For the rod to remain stationary, the magnetic force component along the plane must balance the gravitational component:

\( I B \cos30^\circ = 2.5\,\text{N/m} \)

Solving for \( I \):

\( I = \frac{2.5}{B \cos30^\circ} \)

Substitute \( B = 0.25\,\text{T} \) and \( \cos30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \):

\( I = \frac{2.5}{0.25 \times 0.866} = \frac{2.5}{0.2165} \approx 11.54\,\text{A} \)

The closest answer among the options is (d) \( 11.32\,\text{A} \).

Final Answer: The current flowing in the rod to keep it stationary is approximately \( 11.32\,\text{A} \).