Question

A conducting bar slides without friction on two parallel horizontal rails that are 50 cm apart and connected by a wire at oneend. The resistance of the bar and the rails is constant and equal to 0.10 Ω. A uniform magnetic field is perpendicular to theplane of the rails. A 0.080-N force parallel to the rails is required to keep the bar moving at a constant speed of 0.50 m/s. Whatis the magnitude of the magnetic field?

a. 0.36 Tb. 0.10 Tc. 0.54 Td. 0.93 Te. 0.25 T

Answer 1

Step-by-step explanation:

Given

Distance between rails
`d=50 cm`

Resistance
`R=0.1 \Omega`

Force
`F=0.08 N`

velocity
`v=0.5 m/s`

Magnetic Force is given

`F=B\cdot I\cdot d\cdot \sin \theta`

`F=B*I*d*\sin 90`

Also Emf is given by

`e=B*d*v=I*R`

`I=(B*d*v)/(R)`

thus

`F=(B^2*d^2*v)/(R)`

`B=\sqrt{(F*R)/(d^2v)}`

`B=\sqrt{(0.08*0.1)/(0.5^2*0.5)}`

`B=0.25 T`