Question

A conducting bar slides without friction on two parallel horizontal rails 50 cm apart. The resistance of the bar and the rails is assumed to be constant and equal to 0.1 Ω. A uniform magnetic field is perpendicular to the plane of the rails. A force F = 0.08 N is required to keep the bar moving at a constant speed of 0.5 m/s. What is the magnitude of the magnetic field?

Answer 1

ANSWER:

0.253 T

EXPLANATION:

we know that Lorentz force (force on a moving conductor ) is given by

F= $BILSIN\Theta$

where B= magnetic field

           I= current through conductor

           sinΘ=90° given

so F=BIL

B=$\frac{F}{LI}$

$\frac{dA}{dt}=0.5\times 0.5 =0.25 m^{2}/sec$

$V=\frac{d\left ( BA \right )}{dt}$=B\frac{dA}{dt}

=B×0.25

$I=\frac{V}{0.1}$

=B×2.5

$B^{2}=\frac{0.08}{0.5\times 2.5}$

B=0.253 T