Question

32) 10 A current is passing through a very long wire of radius 5 cm. Then magnetic
field at a distance of 2 cm inside from its curved surface is_________×10^-5 T
(A) 2.4 x 10^-5
(B) 6.7 x 10^-5
(C) 2.4 × 10^5
(D) 2.4​

Answer 1

Given:

10 A current is passing through a very long wire of radius 5 cm.

To find:

magnetic field at a distance of 2 cm inside from its curved surface

Calculation:

Applying Ampere Circutal Law;

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

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

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

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

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

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

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

Putting available values:

$= > \displaystyle \:B = {10}^{ - 7} \times \bigg \{ \frac{2 \times 10}{ {(0.05)}^{2} } \times (0.05 - 0.02) \bigg \}$

$= > \displaystyle \:B = {10}^{ - 7} \times \bigg \{ \frac{6 \times 10}{ {(0.05)}^{2} } \bigg \}$

$= > \displaystyle \:B = {10}^{ - 7} \times \bigg \{ \frac{6 \times 10}{ 25 \times {10}^{ - 4} } \times {10}^{ - 2} \bigg \}$

$= > \displaystyle \:B = \frac{60}{25} \times {10}^{ - 5} \: tesla$

$= > \displaystyle \:B = 2.4 \times {10}^{ - 5} \: tesla$

So, final answer is:

$\boxed{ \sf{ \red{ \displaystyle \:B = 2.4 \times {10}^{ - 5} \: tesla}}}$

Answer 2

Answer:2.4

Explanation: