On a smooth plane inclined at 30 degree with the

Question

On a smooth plane inclined at 30 degree with the horizontal , a thin current – carrying metallic rod is placed parallel to the horizontal ground. The plane is located in a uniform magnetic field of 0.15 T in the vertical direction.For what value of current can the rod remain stationary? The mass per unit length of the rod is 0.03 kg / m.Ans:11.32.Explain with diagram plz.

Arun · Accepted Answer

the current carrying rod will remain stationary when net vector summation all forces acting on the body is zero
hence the force due to magnetic field acting on current carrying rod is Bil in horizontal direction
now taking the component of magnetic force along the plane (up)

F upward = Bi L cos 30 = BiL * sqrt (3) / 2

F downward the gravity = mg sin 30 = mg/2

for equilibrium

F upward = F downward

i = mg / BL * sqrt (3)

i = 0.03 * 10 / 0.15 * 1 * 1.732

i = 1.15 Amp

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Rahul , 7 Years ago

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