Question

A straight horizontal rod of mass 50 gm and length 50 cm is placed in uniform horizontal magnetic field of 0.2T perpendicular to X. Calculate the current in X if the force acting on it just balances its weight.(g = 9.8 m/s^2)

Answer 1

Answer:

I = 4.9A

Explanation:

Please see the attachment below.

Answer 2

Given Information:  

Magnetic field = B = 0.2 T  

Mass of rod = m = 50 g = 0.050 kg  

Length of rod= 50 cm = 0.50 m

Gravitational acceleration = g = 9.8 m/s²

Required Information:  

Current flowing through rod = I = ?

Answer:  

Current flowing through rod = 4.9 A

Explanation:  

The magnetic force produced in a current carrying rod with magnetic field B is given by

F = BILsin(θ)  

This magnetic force is balanced by the weight of the rod which is given by

F = mg

mg = BILsin(θ)

I = mg/BLsin(θ)

Where m is the mass of the rod, L is the length of the rod, B is the magnetic field induced and θ is the angle between magnetic field and current.

I = 0.050*9.8/0.2*0.50*sin(90°)

I = 4.9 A

Therefore, a current of 4.9 A will flow through the rod placed in uniform magnetic field when the force acting on it just balances its weight.