Please solve 3 & 4 A conducting rod moving perpendicularly to a magnetic field w
ID: 2251740 • Letter: P
Question
Please solve 3 & 4
A conducting rod moving perpendicularly to a magnetic field will have a voltage produced between the two ends due to the magnetic force on the electrons being equal to the electrical force inside the wire due to the voltage difference. Suppose a 3 m conductor is moving at 70 miles/hr perpendicularly to a field of 0.5 times 10-4 T. What is the voltage between the ends of the rod? What is the electic field inside the rod? Suppose there is a resistor of 1.5 times 106 Ohms in series with a 1.5 H inductor and a 12 volt battery and a switch. What is the voltage across the inductor one microsec after the switch is closed?Explanation / Answer
3)
a)
voltage between the rods is given by:
V = Blv
where B = magnetic field = 0.5*10^-4 T
l = length of rod = 3m
v =velocity of rod = 70 miles/h = 31.29 m/s
So, V = 0.5*10^-4*3*31.29 = 4.69*10^-3 V <------------answer
b)
electric field inside the rod is given by:
E = V/l = 4.69*10^-3/3
= 1.56*10^-3 V/m <-------------answer
4)
R = 1.5*10^6
L = 1.5 H
E = 12 V
Voltage across the inductor is given by:
V = E*(e^(-t*R/L))
So, at t = 1 us = 1*10^-6 s ,
V = 12*(e^(-1*10^-6*1.5*10^6/1.5))
= 4.41 V <------------answer