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

Question

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^ 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.

Accepted Answer

For the rod to remain stationary on the smooth incline:Component of gravity down the incline

= m g sin θ

.Magnetic force on the rod

F_{m} = I lB

.Since

B

is vertical, the horizontal magnetic force is

I lB

directed such that its component up the incline balances gravity.The force balance along the incline:

I lB cos θ = m g sin θ

.

I = \frac{m}{l} \frac{g tan θ}{B}

Substitute the values:

\frac{m}{l} = 0.5 kg/m, θ = 3 0^{\circ}, B = 0.25 T

.

I = \frac{0.5 \times 9.8 \times tan 3 0 ^{\circ}}{0.25}

I = 2 \times 9.8 \times \frac{1}{3} = \frac{19.6}{1.732} \approx 11.32 A

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