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Two parallel fixed conducting rails are ...

Two parallel fixed conducting rails are `l` distance apart. They are connected by a conducting wire xy. A uniform magnetic field B is applied normal to its plane. Two other rods PQ and RT(RS = ST) are moved with a constant velocity v. Resistance/ length of these two rods are `lamda` no other wire has resistance. Then :

A

current in the conducting wire xy is `(2vB)/(lamda)`

B

current in the conducting wire xy is `(3)/(2)(vB)/(lamda)`

C

To maintain constant velocity of rod RST force required in forward direction is `(B^(2) l v)/(2 lamda)`

D

To maintain constant velocity of rod RST force required in forward direction is `(B^(2) l v)/(lamda)`

Text Solution

Verified by Experts

The correct Answer is:
B, C

Current in PQ wire is `(VB lamda)/(2 lamda) = (VB)/(lamda)`
Current in RST wire `(i_(2)) = (VB lamda)/(2 lamda l) = (VB)/(2 lamda)`
Current in xy wire `= i_(1)+i_(2) = (3)/(2) (V_(B))/(lamda)`
For required to maintain the velocity of RST wire constant
`F=i lamda_(eq)B=((VB)/(2 lamda))(lamda)B-(B^(2)lamdav)/(2 lamda)`.
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