InterviewSolution
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InterviewSolution
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Show that the magnetic field along the axis of a current carrying coil of radius r at a distance x from the centre of the coil is smaller by the fraction `3x^2//2r^2` than the field at the centre of the coil carrying current. |
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Answer» The magnetic field induction, at the centre of circular coil of n turns, radius r, carrying current I is `B_0=(mu_0)/(4pi)(2pinI)/(r)=(mu_0nI)/(2r)` Magnetic field induction at a point on the axis of the circular coil carrying current is given by `B_0=(mu_0)/(4pi)(2pinIr^2)/((r^2+x^2)^(3//2))=(mu_0nIr^2)/(2(r^2+x^2)^(3//2))` `:. B/B_0=(r^3)/((r^2+x^2)^(3//2))=r^3/r^3[1+x^2/r^2]^(-3//2)=[1-3/2x^2/r^2+...]=[1-3/2x^2/r^2]` Fractional decrease in magnetic field `=(B_0-B)/(B_0)=1-B/B_0=1-[1-3/2x^2/r^2]=(3x^2)/(2r^2)` |
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