Question

a.) A 13.0 m long, thin, uniform aluminum pole slides south at a speed of 21.0 m/s. The length of the pole maintains an east-west orientation while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 32.0 µT. What is the magnitude of the induced emf between the ends of the pole?

b.) The west end of the pole impacts and sticks to a pylon, causing the pole to rotate clockwise as viewed from above. While the pole rotates, what is the magnitude of the induced emf between the ends of the pole? (Hint: use conservation of angular momentum to find the speed of the pole after the collision.)

Answer 1

Sol (a) Length L = 13.0 m, Magnetic Field B = 32.0 µT, Velocity V = 21.0 m/s

Magnetic Field B (Vertical),Length L (East west ) and Velocity V (Southwards) are mutually perpendicular.

So Induced Emf = B L V

                             = 32x10-6. 13. 21

                             = 8.7 mV

Sol (b) The rod is moving southwards with speed 21 m/s then it's west end O sticks with a pylon. Now the body will rotate about point O.

Angular momentum of body will be conserved about point O as there isn't any external torque about point ).

l = distance of O from center of mass of rod OC=L/2

Angular momentum of body about point O ( before point O sticks to pylon)

Lintial = M V l = M V (L /2)

Angular momentum of rod with respect to O, Io = ML2/3

w = Angular velocity of rod about point O

Angular momentum of body about point O ( after point O sticks to pylon)

Lfinal = Io. w

= (ML2/3). w

Applying conservation of momentum of body about point O

Lintial = Lfinal

M V (L /2) = (ML2​​​​​​​/3). w

w = (3V/2L)

Now Induced EMF in rotational motion = (1/2)B w L2 = (1/2) B (3V/2L) L2

= (3/4) B V L

= (3/4). 8.7 mV

= 6.5 mV