Question

7. a) A long, straight wire of diameter 5.0 mm carries a uniformly distributed current of
15 A. At what distance from the axis of the wire will the magnitude of B be
maximum? Justify your answer.
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Answer 1

Given :

Diameter of the straight wire :

= 5 mm

Magnitude of uniformly distributed current carried by the wire :

I = 15 A

To Find :

The distance from the  axis of the wire where the magnitude of magnetic field (B) is maximum = ?

Solution:

∴The formula for the magnetic field at any point due to a current carrying is given as :

$B = \frac{\mu i}{2 \pi d}$

Here $\mu=$ permitivity of medium

        i = current

        d = distance of point from the axis of wire

Now consider a point P at a distance d from axis inside  the wire :

∴ $B = \frac {\mu i }{2 \pi d}$

But here i = current enclosed in the loop formed by the circle of radius d.

So, $i = \frac{I}{\pi r^{2} }\times \pi d^{2}$

Here , I = total current through wire

And , r = radius of wire

⇒$B = \frac{\mu }{2 \pi d}\times \frac{I}{\pi r^{2} } \times \pi d^{2}$

     = $\frac{\mu I d }{2\pi r^{2} }$

So B outside the wire will be = $\frac{\mu I}{2 \pi d}$

∴Since B is maximum at the boundary of wire .

So it is maximum at 2.5 mm from the axis.