Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A metallic rod of mass per unit length of 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction of 0.25 T is acting on it in the vertical direction. What is the current flowing through the rod to keep it stationary?

7.14 A

5.98 A

14.76 A

11.32 A

Step-by-Step Solution

Force Analysis: The rod is in equilibrium on the inclined plane under the influence of three forces:

Gravity ( $mg$ ) acting vertically downwards.
Normal reaction ( $N$ ) perpendicular to the plane.
Magnetic force ( $F_m$ ) acting horizontally. Since the magnetic field $\vec{B}$ is vertical and the current $\vec{l}$ is horizontal, the force $\vec{F} = I(\vec{l} \times \vec{B})$ is horizontal . The magnitude is $F_m = IlB\sin(90^\circ) = IlB$ .

Equilibrium Condition: For the rod to remain stationary, the component of forces down the incline must balance the component up the incline.

Component of weight down the incline: $mg \sin \theta$ .
Component of magnetic force up the incline: Since $F_m$ is horizontal, its component along the incline is $F_m \cos \theta$ .

Equation: $mg \sin \theta = F_m \cos \theta$ $mg \sin \theta = (IlB) \cos \theta$
Calculation: Rearranging for current $I$ : $I = \frac{mg}{lB} \frac{\sin \theta}{\cos \theta} = \frac{m}{l} \cdot \frac{g}{B} \tan \theta$ Given: $\frac{m}{l} = 0.5 \text{ kg m}^{-1}$ , $g = 9.8 \text{ m s}^{-2}$ , $B = 0.25 \text{ T}$ , $\theta = 30^\circ$ . $I = \frac{0.5 \times 9.8}{0.25} \times \tan 30^\circ$ $I = \frac{4.9}{0.25} \times 0.5774$ $I = 19.6 \times 0.5774 \approx 11.32 \text{ A}$

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