On a smooth inclined plane at 30 degrees with the horizontal , a thin current carrying metallic rod is placed parallel to the horizontal groundc.The plane is located in a uniform magnetic field of 0.15T in the vertical direction . For what value of current can the rod remain stationery? The mass per unit length of the rod is 0.03kg/m

Answers (1)

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 A

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