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

Answer:

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

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

$I=\frac{B*d*v}{R}$

thus

$F=\frac{B^2*d^2*v}{R}$

$B=\sqrt{\frac{F*R}{d^2v}}$

$B=\sqrt{\frac{0.08*0.1}{0.5^2*0.5}}$

$B=0.25 T$