1. Forces on the rod

The rod lies horizontally on a smooth incline of angle $45^\circ$.

Since the plane is smooth, the rod can remain at rest only if the net force along the incline is zero.

• Weight per unit length of rod: $\lambda g = 0.45 \times 10 = 4.5\ \text{N m}^{-1}$
• Component of weight down the incline: $f_g = \lambda g \sin 45^\circ = 4.5\cdot \frac{1}{\sqrt 2}$

2. Magnetic force on the rod

The rod is horizontal, and the magnetic field is vertically upward. So the angle between current direction and magnetic field is $90^\circ$.

Hence magnetic force per unit length is $f_B = I B$

Its direction is horizontal and perpendicular to the rod. The component of this force along the incline is $f_B \cos 45^\circ = IB \cdot \frac{1}{\sqrt 2}$

3. Condition for minimum current

To just keep the rod stationary, magnetic force component up the incline must balance the component of weight down the incline: $IB\frac{1}{\sqrt 2} = \lambda g \frac{1}{\sqrt 2}$

Cancelling $\frac{1}{\sqrt 2}$, $IB = \lambda g$

Therefore, $I = \frac{\lambda g}{B} = \frac{0.45\times 10}{0.15} = \frac{4.5}{0.15} = 30\ \text{A}$

4. Option check

• A: $30\,\text{A}$ ✔️
• B: $15\,\text{A}$ ✘
• C: $10\,\text{A}$ ✘
• D: $3\,\text{A}$ ✘

Therefore, the correct answer is A.