A conducting bar slides without friction on two parallel horizontal rails that are 50 cm apart and connected by a wire at one end.The resistance of the bar and the rails is constant and equal to 0.10 Ω.A uniform magnetic field is perpendicular to the plane 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.What is the magnitude of the magnetic field?
A) 0.10 T
B) 0.25 T
C) 0.36 T
D) 0.54 T
E) 0.93 T

Question

Quizplus · Accepted Answer

To find the magnitude of the magnetic field, we use the formula for the force on a current-carrying conductor in a magnetic field, $F = BIL$, where $F$ is the force, $B$ is the magnetic field strength, $I$ is the current, and $L$ is the length of the conductor (which in this case is the distance between the rails, 0.50 m). The force required to keep the bar moving at a constant speed is given as 0.080 N, which is equal to the magnetic force. The current $I$ can be found using Ohm's law, $V = IR$, where $V$ is the voltage and $R$ is the resistance. The voltage is generated by the motion of the bar in the magnetic field and can be calculated using $V = BLv$, where $v$ is the velocity of the bar. Substituting $V = BLv$ into $V = IR$ gives $BLv = IR$, and substituting this into $F = BIL$ gives $F = B^2L^2v/R$. Solving for $B$ gives $B = \sqrt{FR/(L^2v)}$. Substituting $F = 0.080 N$, $R = 0.10 Ω$, $L = 0.50 m$, and $v = 0.50 m/s$ gives $B = \sqrt{(0.080 N)(0.10 Ω)/(0.50 m)^2(0.50 m/s)} = 0.25 T$.

A Conducting Bar Slides Without Friction on Two Parallel Horizontal