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A metallic square loop ABCD is moving in...

A metallic square loop `ABCD` is moving in its own plane with velocity `v` in a uniform magnetic field perpendicular to its plane as shown in the figure. Find
`{:a)` In which sides of the loop electric field is induced.
`{:b)` Net emf induced in the loop
`{:c)` If one `'BC'` is outside the field with remaining loop in the field and is being pulled ot a constant velocity then induced current in the loop.

Text Solution

Verified by Experts

`{:a)` The metallic square lop moves in its own plane with velocity `v`.
A uniform magnetic field is imposed perpendicular to the plane of the square loop.
`AD` and `BC` are `_|_` to the velocity as well as `_|_` to field applied. Hence electric field is incduced across the sides `AD` and `BC` only.
`{:b)` As there is no change of flux through the entire coil net emf induced in the coil is zero.
`{: c)` Induced current `i=(e)/( R)` Where `R` is the resistance of the coil.
`implies i=(Blv)/(R )` ( Only the side `AD` cuts the flux)
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Knowledge Check

  • A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced

    A
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    B
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    D
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    A
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    B
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    C
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    D
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