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

Question

contestada

Respuesta :

nuuk nuuk · Answer 1 · 2019-12-18T10:58:58+01:00

Answer:

Explanation:

Given

Distance between rails [tex]d=50 cm[/tex]

Resistance [tex]R=0.1 \Omega [/tex]

Force [tex]F=0.08 N[/tex]

velocity [tex]v=0.5 m/s[/tex]

Magnetic Force is given

[tex]F=B\cdot I\cdot d\cdot \sin \theta [/tex]

[tex]F=B*I*d*\sin 90[/tex]

Also Emf is given by

[tex]e=B*d*v=I*R[/tex]

[tex]I=\frac{B*d*v}{R}[/tex]

thus

[tex]F=\frac{B^2*d^2*v}{R}[/tex]

[tex]B=\sqrt{\frac{F*R}{d^2v}}[/tex]

[tex]B=\sqrt{\frac{0.08*0.1}{0.5^2*0.5}}[/tex]

[tex]B=0.25 T[/tex]