As shown in the figure, a metallic rod of linear density 0.45 kg m$$^{-1}$$ is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be (Use $$g = 10$$ m s$$^{-2}$$)

We need to determine the minimum current ($$I$$) required to keep a metallic rod stationary on a smooth inclined plane under the influence of a vertical magnetic field.

1. Identify the Given Parameters

• Linear mass density ($$\lambda = \frac{m}{l}$$) = $$0.45\text{ kg m}^{-1}$$
• Angle of inclination ($$\theta$$) = $$45^\circ$$
• Magnetic field strength ($$B$$) = $$0.15\text{ T}$$ (directed vertically upwards)
• Acceleration due to gravity ($$g$$) = $$10\text{ m s}^{-2}$$

2. Analyze the Forces Acting on the Rod

For the rod to remain stationary on the smooth inclined plane, the components of the forces acting parallel to the surface of the incline must perfectly balance each other:

• Gravitational Force Component: The component of the rod's weight acting down the incline is $$mg \sin\theta$$.
• Magnetic Force Component: A horizontal magnetic force ($$F_m = I l B$$) acts on the rod due to the vertical magnetic field. Resolving this force parallel to the incline, the component pushing up the incline is $$I l B \cos\theta$$.

3. Set Up the Equilibrium Equation

Equating the force component acting up the incline to the force component acting down the incline:

$$I l B \cos\theta = mg \sin\theta$$

Rearranging the formula to solve for the current ($$I$$):

$$I = \left(\frac{m}{l}\right) \frac{g}{B} \cdot \frac{\sin\theta}{\cos\theta}$$

$$I = \lambda \frac{g}{B} \tan\theta$$

4. Calculate the Minimum Current

Substitute the given numerical values into our derived equilibrium equation:

$$I = 0.45 \times \frac{10}{0.15} \times \tan(45^\circ)$$

Since $$\tan(45^\circ) = 1$$:

$$I = \frac{4.5}{0.15} \times 1 = 30\text{ A}$$

Therefore, the minimum current flowing in the rod required to keep it stationary is 30 A, which corresponds to Option A.