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A square coil of side 30 cm with 500 tur...

A square coil of side 30 cm with 500 turns is kept in a uniform magnetic field of 0.4 T. The plane of the coil is inclined at an angle of `30^(@)` to the field. Calculate the magnetic flux through the coil.

Text Solution

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Given : Side of a square coil `=30cm=30xx10^(-2)m`
Area of a square coil `A=" side "xx" side "`
`=30xx30xx10^(-4)m`
Magnetic field B = o.4 T
The plane niclined to the field `=30^(@)`
`thereforetheta=60^(@)(90-30)i.etheta=" angle between normal surface "`
No. of turnsd N = 500
Formula :
`phi=" NA "Bcostheta`
`phi=9xx10^(-2)xx500xx0.4xxcos60^(@)`
`=9xx10^(-2)xx500xx0.4xx(1)/(2)`
`phi=9Wb`
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