A metallic rod of mass per unit length of 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which ma

Question

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?

Accepted Answer

1. **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 $\vec{F} = I(\vec{l} 	imes \vec{B})$ is horizontal . The magnitude is $F_m = IlB\sin(90^\circ) = IlB$.
2. **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: $mg \sin 	heta$ .
   - Component of magnetic force up the incline: Since $F_m$ is horizontal, its component along the incline is $F_m \cos 	heta$.
3. **Equation:**
   $$mg \sin 	heta = F_m \cos 	heta$$
   $$mg \sin 	heta = (IlB) \cos 	heta$$
4. **Calculation:** Rearranging for current $I$:
   $$I = \frac{mg}{lB} \frac{\sin 	heta}{\cos 	heta} = \frac{m}{l} \cdot \frac{g}{B} 	an 	heta$$
   Given: $\frac{m}{l} = 0.5 	ext{ kg m}^{-1}$, $g = 9.8 	ext{ m s}^{-2}$, $B = 0.25 	ext{ T}$, $	heta = 30^\circ$.
   $$I = \frac{0.5 	imes 9.8}{0.25} 	imes 	an 30^\circ$$
   $$I = \frac{4.9}{0.25} 	imes 0.5774$$
   $$I = 19.6 	imes 0.5774 \approx 11.32 	ext{ A}$$

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