Question

A long straight wire of radius a carries a steady current. The current is uniformly distributed across its cross-section.the ratio of the magnetic field at $\frac{a}{4}$ inside and $\frac{a}{4}$ outside from the surface of wire is:

Answer 1

Solution :

We know,magnetic field B in inside points = $\frac{{\mu }_{0}ir}{2\pi R^2}$ [where,R = radius,r = distance of the inside point from the axis and i = total current]

Distance from the axis at $\frac{a}{4}$ inside = $(a-\frac{a}{4})$ =$\frac{3a}{4}$

So,Magnetic field at $\frac{a}{4}$ inside = $\frac{{\mu }_0\times i\times \frac{3a}{4}}{2\pi a^2}$

= $\frac{3{\mu }_{0}i}{8\pi a}$

Distance from the axis at $\frac{a}{4}$ outside = $(a+\frac{a}{4})$

We also know,magnetic field B at outside points = $\frac{{\mu }_{0}i}{2\pi r}$ [r=distance of the outside point from the axis]

So,magnetic field at $\frac{a}{4}$ outside = $\frac{{\mu }_{0}i}{2\pi \times \frac{5a}{4}}$

= $\frac{2{\mu }_{0}i}{5\pi a}$

The ratio of the magnetic field at $\frac{a}{4}$ inside and $\frac{a}{4}$ outside the surface of the wire = $\frac{\frac{3{\mu }_{0}i}{8\pi a}}{\frac{2{\mu }_{0}i}{5\pi a}}$

=15:16

So,the correct option is C.