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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 - Physics

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प्रश्न

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?

पर्याय

  • BLv

  • BLv2

  • `1/2Blv`

  • `(2Bl)/v`

MCQ

उत्तर

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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पाठ 12: Electromagnetic induction - Exercises [पृष्ठ २८६]

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बालभारती Physics [English] 12 Standard HSC Maharashtra State Board
पाठ 12 Electromagnetic induction
Exercises | Q 1.2 | पृष्ठ २८६

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