The diagram below shows a conducting rod of lenght 35.0 cm that is free to slide
ID: 1914740 • Letter: T
Question
The diagram below shows a conducting rod of lenght 35.0 cm that is free to slide on two parallel conducting bars. Two resistors R1=6.00 ohms R2=4.00ohms and a capacitor C=8.00 mF are connected across the ends of the bars, a switch S is placed in series with the capacitor. A uniform and constant magnetic field B = 2.75T is directed perpendicular to the plane of the circuit, coming out of the page. An external agent pulls the rod to the left at a constant speed v=8.50m/s.
a) Find the current in each resistor.
b) After the rod has been moving for a few instants, what is the charge on the capacitor?
c) What is the total electrical power dissipated in the circuit?
d) What is the magnitude of the force applied on the rod by the external agent?
Explanation / Answer
in equilbrium condition
we have Fb(magnetic force)=Fe(electric force)
qvB= qV/l v--> velocity of the rod, B--> magnetic field V--> potiential diff between rods
l--> length of the rod
V= Bvl
=2.75*35/100*8.50=8.18volts
a) current I= V/R
in 6 resistance I=1.36A,
in4 I=2.045A
b) q=CV
=65.45 *10^(3)C
c) total power = V*Itot= 8.18125*(1.36+2.045)27.86 watt
d) F=BItotl=2.75*(1.36+2.045)*0.35=3.277 N