Problem 1: A horizontal straight wire 5cm long weighing 1.2g/m is placed perpendicular to a uniform horizontal field of 0.6Weber/m². If the resistance of the wire is 3.8 Ohm/m, calculate the potential difference to applied between the ends of the wire to make it just self supporting.

Solution:

Step: 1 Overview

Force on a current carrying conductor in a uniform magnetic field:
Likewise charges, every magnet attracts other magnet, and when current passes through a conductor, conductor creates its own magnetic field. So when a conduct is placed in a uniform magnetic field, it experiences a force which is equal to:

F=BILSinX, where X is an angel between uniform magnetic field and conductor.

Step: 2 Calculation

Given:

L=5cm=0.05m
Mass=1.2g/m
X=90^o
B=0.6W/m²
Resistance=3.8 Ohm/m
Total mass of 5cm(0.05m) wire: m = 0.05m x 1.2g/m = 0.06g = 0.00006Kg
Total resistance of wire: R = 0.05m x 3.8 Ohm/m = 0.19 Ohm.

Required: Voltage at which both gravitational and magnetic force becomes equal and makes the rod self supporting.
Calculating Weight (Gravitational Force):

W = mg
W = 0.00006 x 9.8
W = 0.000588N

Force due to magnetic field:

F=BILSinX
Since I=V/R
F=B(V/R)LSinX
F=VBLSinX/R
VBLSinX=FR
V=FR/(BLSinX)
Since we have to find voltage at which both gravitational and magnetic forces becomes equal, so putting F=W in above equation we get:
V=WR/(BLSinX)
V=(0.000588)(0.19)/((0.6)(0.05)(Sin90^o))
V=0.00011172/0.03
V=0.003724V
V= 3.724 x 10^-3V (Ans)