Question

A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop?

Answer 1

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

$B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }$

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

$B(0) = \frac{\mu_{o} i}{2R}$

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

$\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}$

$2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }$

$4R^{6} = (r^{2} +R^{2} )^{3}$

$r^{2} =0.587R^{2}$

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

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