Question

The magnetic field due to a straight conductor of uniform cross section of radius a and carrying a steady current is represented by

Answer 1

here we can explain by three possible cases.

case 1 : inside the straight conductor :

Take a surface inside the conductor of radius r.

then, current enclosed this surface, i' = i/πa² × πr² = ir²/a²

and length of wire , dl = 2πr

from ampere circuital law,

$\oint{B.dl}=\mu i_{enclosed}$

or, B × 2πr = $\mu$ × ir²/a²

or, B = $\frac{\mu ir}{2\pi a^2}$

here it is clear that B is directly proportional to r. so, graph between B and r  inside the conductor is straight line.

case 2 : surface of conductor :

enclosed current, I = i

length of wire, dl = 2πa

so, B = $\frac{\mu i}{2\pi a}$

at surface of conductor magnetic field intensity will be maximum e.g., $\frac{\mu i}{2\pi a}$

case 3 : outside the conductor :

enclosed current = i

length of wire, dl = 2πr

so, magnetic field, B = $\frac{\mu i}{2\pi r}$

here it is clear that magnetic field is inverse proportional to r outside the conductor. so, graph between B and r outside the conductor will be hyperbolic.

hence, option (a) is correct choice.