Question

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point?

a) A conducting rod is free to slide on two parallel rails with negligible friction. At the right end of the rails, a voltage source of strength V in series with a resistor of resistance R makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. The rails extend to infinity on the left. The arrangement is shown in the figure.b) Assuming that the rails have no resistance, what is the most accurate qualitative description of the motion of the rod?

c) What is the acceleration a() of the rod? Take m to be the mass of the rod. d) What is the terminal velocity ve reached by the rod?

Answer 1

(a)the current in the rod is down because the current in the circuit is in a counter-clockwise direction.

From the equation force on current-carrying wire \(\mathrm{F}=\Pi \times \mathrm{B}\) L represents the direction of the current.

since the force point to the left, the right-hand rule indicates magnetic field points out of the page.

(b) Applying Lenz law to the given rod. The change in flux that

induced this current is caused by the rod's motion caused by the current flowing around the circuit due to the voltage source.

This induced current works against that source current and reduces it alternatively.

the higher velocity of the rod, the higher induced emf and lower current flowing through the \(100 \mathrm{p}\)

But the rod's acceleration is proportional to the current; hence, acceleration goes down as the velocity goes up. The velocity cannot increase beyond the point, which is the induced emf is equal to the opposite of the force.

(c)

The magnetic flux out of the loop circuit is \(\phi_{B}=\mathrm{BLx}(t)\)

The induced \(\operatorname{emf}\) is \(\varepsilon_{i x d}=-\frac{d \phi_{B}}{d t}\)

$$ \begin{array}{l} =-\frac{\mathrm{BLx}(t)}{d t} \\ =-B L \nu(t) \end{array} $$

The net emf in the circuit is

\(\varepsilon_{n e t}=V+\varepsilon_{i x j}\)

$$ =V-B L v(t) $$

The current in the circuit is \(I_{\max }=\frac{\varepsilon_{n e t}}{R}\)

$$ =\frac{V-B L \nu(t)}{R} $$

The force acting on the rod is \(\mathrm{F}(t)=\Pi \mathrm{B}\)

$$ =\frac{V-B L \nu(t)}{R} L B $$

The accleration is \(a(t)=\frac{F(t)}{m}\)

$$ \begin{array}{l} =\frac{\frac{V-B L \nu(t)}{R} L B}{\frac{m}{M}} \\ =\frac{(V-B L \nu(t)) L B}{R m} \end{array} $$

D) V(t) = V / (BL)