A horizontal rod of mass `10g` and length `10cm` is placed on a smooth plane inclined at an angle of `60^@` with the horizontal with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field induction B is applied vertically downwards. If the current through the rod is `1*73ampere`, the value of B for which the rod remains stationary on the inclined plane is

Options

A. `1.73` T
B. `(1)/(1.73)`T
C. 1.T
D. None of the above

Answer

To solve the problem, we need to find the value of the magnetic field induction \( B \) that keeps the rod stationary on the inclined plane. We will use the balance of forces acting on the rod.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Rod:**
- The weight of the rod \( W = mg \) acts vertically downwards.
- The magnetic force \( F_B \) acts perpendicular to the direction of the current and the magnetic field.

2. **Resolve the Weight of the Rod:**
- The weight of the rod can be resolved into two components:
- Perpendicular to the inclined plane: \( W_{\perp} = mg \cos(60^\circ) \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin(60^\circ) \)

3. **Magnetic Force on the Rod:**
- The magnetic force acting on the rod is given by:
\[ F_B = I \cdot L \cdot B \cdot \sin(90^\circ) = I \cdot L \cdot B \]
- Since the rod is horizontal and the magnetic field is vertical, the angle between the current direction and the magnetic field is \( 90^\circ \).

4. **Set Up the Equation for Equilibrium:**
- For the rod to remain stationary, the magnetic force must balance the component of the weight acting down the incline:
\[ F_B = W_{\parallel} \]
- Therefore, we have:
\[ I \cdot L \cdot B = mg \sin(60^\circ) \]

5. **Substitute Known Values:**
- Given:
- Mass \( m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} \)
- Length \( L = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} \)
- Current \( I = 1.73 \, \text{A} \)
- \( g \approx 10 \, \text{m/s}^2 \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

- Plugging in the values:
\[ 1.73 \cdot (10 \times 10^{-2}) \cdot B = (10 \times 10^{-3}) \cdot 10 \cdot \frac{\sqrt{3}}{2} \]

6. **Simplify the Equation:**
- This simplifies to:
\[ 1.73 \cdot 10^{-1} \cdot B = 5 \sqrt{3} \times 10^{-3} \]
- Rearranging gives:
\[ B = \frac{5 \sqrt{3} \times 10^{-3}}{1.73 \times 10^{-1}} \]

7. **Calculate the Value of B:**
- Calculating the right-hand side:
\[ B = \frac{5 \sqrt{3}}{1.73} \times 10^{-2} \]
- Approximating \( \sqrt{3} \approx 1.732 \):
\[ B \approx \frac{5 \times 1.732}{1.73} \times 10^{-2} \approx 5 \times 10^{-2} \approx 0.05 \, \text{T} \]

8. **Final Result:**
- Thus, the value of \( B \) for which the rod remains stationary on the inclined plane is approximately \( 1 \, \text{T} \).