1.

Find the distance at which the magnetic field on axis as compared to the magnetic field at the centre of the coil carrying current I and radius R is 1/8.

Answer»

R
`sqrt2R`
2R
`sqrt3R`

Solution :`B_("centre")/B_("axis")=[1+x^(2)/R^(2)]^(3/2),"But "B_("axis")=(B_("centre"))/8`
`thereforeB_("centre")/B_("axis")XX8=[1+x^(2)/R^(2)]^(3/2)`
`THEREFORE(2^(3))^(2/3)=1+x^(2)/R^(2)`
`therefore(2^(2))=[1+x^(2)/R^(2)]`
`therefore4=1+x^(2)/R^(2)" "thereforex^(2)/R^(2)=3rArrx=sqrt3R`


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