A metallic rod of mass per unit length `0.5 kg m^(-1)` is lying horizontally on a straght inclined plane which makes an angle of `30^(@)` with the horizontal. The rod is not allowed to slide down by flowing a current throguh 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

Options

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

Explanation

To solve the problem step by step, we will analyze the forces acting on the metallic rod and apply the necessary physics principles.

### Step 1: Identify the forces acting on the rod
The rod is subjected to two main forces:
1. The gravitational force acting downwards, \( F_g = mg \).
2. The magnetic force acting perpendicular to the current and magnetic field, \( F_m = IBL \).

### Step 2: Calculate the gravitational force component along the incline
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( F_{g\perp} = mg \cos(\theta) \)
- Parallel to the incline (down the slope): \( F_{g\parallel} = mg \sin(\theta) \)

### Step 3: Set up the equilibrium condition
For the rod to remain stationary on the incline, the magnetic force must balance the gravitational force component acting down the slope:
\[ F_m = F_{g\parallel} \]
Thus,
\[ IBL = mg \sin(\theta) \]

### Step 4: Substitute known values
Given:
- Mass per unit length, \( \frac{m}{L} = 0.5 \, \text{kg/m} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
- Angle of incline, \( \theta = 30^\circ \)
- Magnetic field, \( B = 0.25 \, \text{T} \)

From the mass per unit length, we can express \( m \) as:
\[ m = \frac{m}{L} \cdot L = 0.5L \]

Now substituting into the equilibrium equation:
\[ IBL = (0.5L)(9.8) \sin(30^\circ) \]

### Step 5: Simplify the equation
Since \( \sin(30^\circ) = 0.5 \):
\[ IBL = (0.5L)(9.8)(0.5) \]
\[ IBL = 0.25L \cdot 9.8 \]
\[ IBL = 2.45L \]

### Step 6: Cancel \( L \) from both sides
Assuming \( L \) is not zero, we can cancel \( L \):
\[ IB = 2.45 \]

### Step 7: Solve for current \( I \)
Now we can solve for \( I \):
\[ I = \frac{2.45}{B} \]
Substituting \( B = 0.25 \, \text{T} \):
\[ I = \frac{2.45}{0.25} \]
\[ I = 9.8 \, \text{A} \]

### Step 8: Final calculation
To find the current, we can also use the mass per unit length directly:
\[ I = \frac{m/L \cdot g \cdot \sin(30^\circ)}{B} \]
Substituting the values:
\[ I = \frac{0.5 \cdot 9.8 \cdot 0.5}{0.25} \]
\[ I = \frac{2.45}{0.25} \]
\[ I = 9.8 \, \text{A} \]

### Final Answer
The current flowing in the rod to keep it stationary is:
\[ I = 9.8 \, \text{A} \]