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Question
Choose the correct option:
A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
Options
BLv
BLv2
`1/2Blv`
`(2Bl)/v`
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Solution
BLv
Explanation:
Step 1:
Let F = Force on charge q due to the motion of rod in the field
Therefore, `F = qVB` .....(1)
Where, q = charge on body (i.e. rod)
v = velocity of rod
B = Magnetic Field on the rod
Step 2:
This force on the charge is attributed to the induced electric field E
Therefore, `Eq = F` ......(2)
Substitute the values from (1) and (2) and calculate:
`E = V B` .....(3)
Therefore, an electric field `E = V B` is said to be induced in the rod due to the motion in the magnetic field.
Potential difference due to the field = emf induced (e) = Electric Field x Length
= `El`
Using (3),
e = `V Bl`
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