CBSE Class 12 · 2010 · Magnetic Effects of CurrentShort Answer

A long straight wire of circular cross-section of radius

a

carries a steady current

I

distributed uniformly across the cross-section. Apply Ampere's circuital law to calculate the magnetic field at a point

r

for
(i)

r < a

and
(ii)

r > a

Answer: (i) $B = \frac{μ _{0} I r}{2 π a ^{2}}$ for $r < a$ ;
(ii) $B = \frac{μ _{0} I}{2 π r}$ for $r > a$ .

Solution

Take a circular amperian loop of radius

r

coaxial with the wire; by symmetry

B

is tangential and uniform, so

\oint B \cdot d ℓ = B (2 π r)

. (i) For

r < a

the enclosed current is the fraction

I^{'} = \frac{I}{π a ^{2}} π r^{2} = \frac{I r ^{2}}{a ^{2}}

, so

B (2 π r) = μ_{0} \frac{I r ^{2}}{a ^{2}} \Rightarrow B = \frac{μ _{0} I r}{2 π a ^{2}} (B \propto r) .

(ii) For

r > a

the whole current

I

is enclosed, so

B (2 π r) = μ_{0} I \Rightarrow B = \frac{μ _{0} I}{2 π r} (B \propto \frac{1}{r}) .

The field rises linearly inside and falls as

1/ r

outside, matching at the surface

r = a

CBSE Class 12 Board · 2010 · Subjective