Problem 21.21 A very thin 22.0 cm copper bar is aligned horizontally along the e

Question

Problem 21.21

A very thin 22.0 cm copper bar is aligned horizontally along the east-west direction.

Part A

If it moves horizontally from south to north at 15.5 m/s in a vertically upward magnetic field of 1.26 T , what potential difference is induced across its ends ?

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Part B

If it moves horizontally from south to north at 15.5 m/s in a vertically upward magnetic field of 1.26 T , which end (east or west) is at a higher potential ?

Enter your answer as "east" or "west".

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Part C

What would be the potential difference if the bar moved from east to west instead?

Problem 21.21

A very thin 22.0 cm copper bar is aligned horizontally along the east-west direction.

Part A

If it moves horizontally from south to north at 15.5 m/s in a vertically upward magnetic field of 1.26 T , what potential difference is induced across its ends ?

Vab =   V  

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Part B

If it moves horizontally from south to north at 15.5 m/s in a vertically upward magnetic field of 1.26 T , which end (east or west) is at a higher potential ?

Enter your answer as "east" or "west".

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Part C

What would be the potential difference if the bar moved from east to west instead?

Vab =   V  

Explanation / Answer

Part A
Emf Induced is given by, E = B*v*L
where,
B = 1.26 T
v = 15.5 m/s
L = 22 cm = 0.22 m

Emf Induced = 1.26 * 15.5*0.22 V
Emf Induced, E = 4.3 Volt

Part B-
As the Force will be towards East, Therefore Positive Charge will move towards East Direction and Neagtive charge will move towards West Direction.

Thus East side will be at Higher Potential.

Part C-
Force on the Electron will be in North Direction therefore Potential difference between the ends of rod, Vab = 0
Vab = 0

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