Question

An infinitely long straight wire contains a uniformly
continuous current of 10A. The radius of the wire is 4× 10 ⁻²
m. The magnetic field at 2 × 10⁻² m from the centre of the
wire will be:
(a) 0 (b) 2.5 × 10 ⁻⁵ T
(c) 5.0 × 10 ⁻⁵ T (d) none of these.

Answer 1

Given:

An infinitely long straight wire contains a uniformly continuous current of 10A. The radius of the wire is 4× 10 ⁻² m.

To find:

The magnetic field at 2 × 10⁻² m from the centre.

Calculation:

Applying Ampere Circutal Law ;

$\therefore \: \displaystyle \int \: B \: . \: dl = \mu_{0}i$

$= > \: \: B \times 2\pi r = \mu_{0} \times ( \dfrac{I}{\pi {R}^{2}h } \times \pi {r}^{2} h)$

$= > \: \: B \times 2\pi r = \mu_{0} \times ( \dfrac{I {r}^{2} }{ {R}^{2} } )$

$= > \: \: B \times 2\pi = \mu_{0} \times ( \dfrac{I r}{ {R}^{2} } )$

$= > \: \: B = \dfrac{\mu_{0} }{2\pi}\times ( \dfrac{I r}{ {R}^{2} } )$

$= > \: \: B = \dfrac{\mu_{0} }{4\pi}\times ( \dfrac{2I r}{ {R}^{2} } )$

Putting the available values:

$= > \: \: B = {10}^{ - 7} \times \{ \dfrac{2 \times 10 \times 2 \times {10}^{ - 2} }{ {(4 \times {10}^{ - 2} )}^{2} } \}$

$= > \: \: B = {10}^{ - 7} \times \{ \dfrac{2 \times 10 \times 2 \times {10}^{ - 2} }{16 \times {10}^{ - 4} } \}$

$= > \: \: B = {10}^{ - 7} \times \{ \dfrac{ {10}^{ - 1} }{4 \times {10}^{ - 4} } \}$

$= > \: \: B = 2.5 \times {10}^{ - 5} \: tesla$

So, final answer is

$\boxed{ \bold{ \large{\: \: B = 2.5 \times {10}^{ - 5} \: tesla}}}$