Very thin 15.0 cm copper bar is aligned horizontally a l ong the east - west direction. If it moves horizontally from south to north at 11.5 m/s in a vertically upward magnetic field of 1.22 T, A) what potential difference is induced across its ends, and B) which end (east or west) is at a higher potential

2.10 V

Explanation:

To find the potential difference across the ends of the bar, you take into account that when the bar moves with speed v, in a constant magnetic field, the charges in the wire feels a magnetic force that separate the opposite charges, generating an induced potential difference given by the following formula:

  (1)

v: speed of the bar = 11.5 m/s

B: magnitude of the magnetic field = 1.22 T

L: length of the bar = 15.0cm = 0.15m

You replace the values of v, B and L in the equation (1):

hence, the induced potential difference is 2.10 V

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j