A 40 g, 50 cm rod on smooth rails inclined at 45° in a vertical 0.2 T field. Find the current for equilibrium — Physics problems | PhysLabo
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Equilibrium of a current-carrying rod on inclined rails

Medium 11 grade Electricity / Magnetism
Problem statement

A horizontal rod of mass $m = 40$ g and length $L = 50$ cm lies on two smooth parallel rails inclined at an angle $\alpha = 45°$ to the horizontal. The system is in a uniform magnetic field with a vertical induction vector $B = 0.2$ T (the figure is a side view, the rod is shown as a circle). What current must flow through the rod to keep it in equilibrium, and in which direction: away from the observer into the plane of the figure, or toward the observer? Take $g = 10$ m/s².

Side view: a rod on smooth rails inclined at angle α in a vertical magnetic field B; mg, N and the Ampère force are shown
Given
  • $m = 40$ g $= 0.04$ kg
  • $L = 50$ cm $= 0.5$ m
  • $B = 0.2$ T
  • $\alpha = 45°$
  • $g = 10$ m/s²
  • $I = ?$
Hints
Hint 1 / 3
The rod is perpendicular to $\vec B$, so $F_A = BIL$. With a vertical $\vec B$ the Ampère force is horizontal.
Hint 2 / 3
The three forces — $m\vec g$, $\vec N$ and $\vec F_A$ — balance. Project them onto the axis along the rails: $N$ drops out and you get $F_A\cos\alpha = mg\sin\alpha$.
Hint 3 / 3
To keep the rod from sliding down, the Ampère force must point horizontally toward the slope. Find the current direction with the left-hand rule.
Solution

The forces on the rod are gravity $m\vec g$, the normal force $\vec N$ from the rails (perpendicular to the rails — there is no friction) and the Ampère force $\vec F_A$. The rod is perpendicular to $\vec B$ and $\vec B$ is vertical, so the Ampère force is horizontal with magnitude $F_A = BIL$.

Projections of the forces onto the axis along the rails
Projections of the forces onto the axis along the rails

1. Equilibrium. Project the forces onto the axis along the rails ($N$ does not enter the equation):

$$F_A\cos\alpha - mg\sin\alpha = 0 \;\Rightarrow\; F_A = mg\tan\alpha$$

2. Substitute $F_A = BIL$:

$$I = \frac{mg\tan\alpha}{BL} = \frac{0.04 \cdot 10 \cdot 1}{0.2 \cdot 0.5} = 4 \text{ A}$$

3. Direction. To hold the rod, the Ampère force must point horizontally toward the slope (to the right in the figure). By the left-hand rule (the lines of $\vec B$ enter the palm, the thumb points along the force) the fingers must point away from the observer, into the plane of the figure.

Answer

$I = 4$ A; the current flows away from the observer, into the plane of the figure

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