Physics · Moving Charges And Magnetism · NEET
The magnetic field B is vertical (straight up) and the current I flows horizontally along the rod. Force F = I(L x B). Since L is horizontal and B is vertical, F = BIl points horizontally (into the hill), perpendicular to both. It is NOT along the slope. So you must take its component along the slope, which gives the cos(theta) factor.
Balance ALONG the slope. On a smooth incline the normal force is unknown and perpendicular to the surface, so resolving along the slope makes the normal force drop out. Down-slope: mg sin(theta). Up-slope component of the horizontal force: BIl cos(theta). Set them equal: mg sin(theta) = BIl cos(theta).
For the balance equation, no. Smooth means no friction, so the only forces are weight (mg, down), normal N (perpendicular to surface), and magnetic force (horizontal). Resolving along the slope removes N entirely, so you never need its value to find the current.
From dividing the two slope equations. mg sin(theta) = BIl cos(theta). Divide both sides by cos(theta): mg tan(theta) = BIl. So I = mg tan(theta)/(Bl) = (m/l)g tan(theta)/B. The tan appears because gravity uses sin and the horizontal force uses cos.
It depends on the direction of B. If B is set so the magnetic force acts UP the slope (parallel to the surface), then BIl = mg sin(theta) directly, giving I = (m/l)g sin(theta)/B. The tan form is only for a VERTICAL B (horizontal force). Always draw B first, then decide.
A metallic rod of mass per unit length 0.5 kg/m is lying horizontally on a smooth inclined plane which makes an angle of 30 degrees with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
I = (m/l) g tan(theta) / B, where m/l is mass per unit length, theta is the incline angle, and B is the vertical magnetic field. It comes from mg sin(theta) = BIl cos(theta).
Because we resolve forces along the slope. The normal force is perpendicular to the slope, so its slope-component is zero and it cancels out. That is why a smooth incline problem needs no value of N.
The current must flow so that F = I(L x B) points horizontally INTO the incline (toward the hill). This gives an up-slope component that opposes gravity. Reverse the current and the rod slides down faster.
Then the geometry changes. Redraw the force using F = I(L x B) and resolve along the slope again. The sin/cos factors and the final formula depend entirely on the field direction, so always sketch B first.
On a smooth (frictionless) incline, no. If the problem says rough, add the friction force mu*N along the slope and it appears in the balance equation, making the algebra longer.
