Question

A conducting rod whose length is l = 30 cm is placed on a U-shaped metal wire that is connected to a light bulb having a resistance of 5.0 s as shown in the figure. The wire and the rod are in the plane of the page. A constant uniform magnetic field of strength 0.50 T is applied perpendicular to and out of the paper. An external applied force moves the rod to the left with a constant speed of 12 m/s. 

a) What are the magnitude and direction of the induced current in the circuit? 

b) What is the magnitude of the external force required to maintain the constant speed? 

Answer 1

emf of moving rod in magnetic field is given by,

emf = B*v*L

here, B = magnetic field = 0.50 T

v = speed of rod = 12 m/s

L = length of rod = 30 cm = 0.30 m

So,

emf = 0.50*12*0.30

Using ohm's law,

emf = I*R

given, R = resistance of circuit = 5.0 ohm

then,

I = current in circuit = emf/R = B*v*L/R

I = 0.50*12*0.30/5.0 = 0.36 A

So, Magnitude of induced current = 0.36 A

Now direction of induced current will be given by right hand rule:

According to this if we point our thumb towards the direction of motion of charged particle and fingers towards the direction of magnetic field than direction of palm will be direction of force on the positive charge particle.

Now in given case if we point our fingers in the direction of magnetic field (Out of the page) and our thumb towards the direction of motion (toward left), then our palm will be in upward direction, So force on positive charge in the rod will be upward, and force on negative charge will be downward. Now direction of current is in opposite direction of motion of electrons, So Current will be moving in upward direction.

Direction of current = Upward (Counter-clockwise)

now, by force balance on rod,

F_applied - F_magnetic = m*a

given, rod is moving with constant speed.

then, a = 0

So,

F_applied = F_magnetic = force on rod due to magnetic field = I*L*B

F_applied = (B*v*L/R)*L*B = B^2*v*L^2/R

F_applied = (0.50^2*12*0.30^2/5.0) = 0.054

F_applied = 5.4*10^-2 N

Let me know if you've any query.