Consider a straight piece of length x of a wire carrying a current i.Let P be a point on the perpendicular bisector of the piece,situated at a distance d from its middle point.Show that for d>>x,the magnetic field at P varies as 1/ $d^{2}$ whereas d<<x,it varies as 1/d.

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Solution

Since O is the mid-point of MN=l

So, $B = \frac{μ_{0} i}{4 π d} . \frac{2 l}{\sqrt{l^{2} + 4 d^{2}}}$

If $d >> x$ ,thus

$B = \frac{μ_{0} i}{4 π d} . \frac{2 l}{2 d}$

i.e., $B \propto \frac{1}{d^{2}}$

If $d << x$ , then

$B = \frac{μ_{0} i}{4 π d} . \frac{2 l}{l}$

i.e., $B \propto \frac{1}{d}$

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Q. Consider a straight piece of length x of a wire carrying a current i. Let P be a point on the perpendicular bisector of the piece, situated at a distance d from its middle point. Show that for d >> x, the magnetic field at P varies as 1/d² whereas for d << x, it varies as 1/d.

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