A conducting rod of mass m = 10 g and electrical resistance of R_r = 1 Ohm lies

Question

A conducting rod of mass m = 10 g and electrical resistance of R_r = 1 Ohm lies across two parallel perfectly conducting rails in the direction of the x-axis separated by a distance L = 10 cm with the rod perpendicular to the rails along the y-axis. The rails are joined into an electrical circuit with a switch, a resistor R = 10 M Ohm and a capacitor C = 6 mu F, all in a horizontal plane, see the figure. The rest of the circuit has zero resistance. Additionally there is a uniform vertical magnetic field B = 1 T as shown (note the directions of everything). At time t = 0 the capacitor is charged with charge Q_O = 1.67 times 10^-3 C and the rod is at rest at x(0) = 0 when the switch is closed. Calculate the speed v(t) of the rod in terms of the given quantities.

Explanation / Answer

CHarge on capacitor = Qo = 1.67*10^-3 C
Capacitor discharge equation for voltage across the capacitor = V(t) = Vo*exp(-t/RC)
where Vo is the initial voltage, Vo = Qo/C

Now, if current in the circuit is i
force on the rod = Bil [ l is length of the rod, B is the magnetic field]
acceleration of the rod, a = F/m = Bil/m
v(t) = at = Bilt/m

now, i = V(t)/R
R = Rr + R = 1 + 10,000,000 = 10,000,001 ohm
v(t) = Blt*Vo*exp(-t/RC)/10,000,001 *m
m = 0.01 kg
R = 10,000,001 ohm
C = 6*10^-6 F
B = 1 T
l = 0.1 m
Vo = Qo/C = 1.67*10^-3 / 6*10^-6

hence v(t) = 2.7833*10^-4t *exp(-t/60.00006)

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