Question

Answer this question fast !!!​

Answer 1

Explanation:

Answer :

we know that,

$\dashrightarrow \bf \: \vec{B} = \frac{ \mu_0I}{2\pi \: a} \hat {i} \\ \\ \\ \dashrightarrow \bf \: \vec{B} = \frac{ \mu_0}{2\pi } \: \: \frac{ \lambda \: v }{a} \hat {i}$

$\underline{ \boxed{ \red{ \tt{ \lambda \: = \frac{q}{l} \: \: and \: \:I \: = \frac{q}{t} = \frac{ \lambda \: l}{t} = \lambda \: v}}}} \bigstar$

$\\ \\ \\ \dashrightarrow \bf \: \vec{E} \: </u><u>= \frac{ \lambda \hat {j}}{2\pi \epsilon_0a}</u><u> \\ \\ \\ as \: \: we \: know \: that \: \\ \\ \dashrightarrow \bf \: \vec{S} \: = \frac{1}{ \mu_0} (\vec{E} \times \vec{B})$

$\\ \\ \\ \dashrightarrow \bf \: \vec{E} \: = \</u><u>frac{ \lambda \hat {j}}{2\pi \epsilon_0a} \\ \\ \\ as \: \: we \: know \: that \: \\ \\ \dashrightarrow \bf \: \vec{S} \: = \frac{1}{ \mu_0} (\vec{E} \times \vec{B})$

$\dashrightarrow \bf \: \vec{S} = \frac{1}{ \mu_0 } \bigg[ \: \frac{ \lambda \: }{2\pi \epsilon_0a} \hat{j} \times \frac{ \mu_0}{2\pi \: a} \lambda \: v \hat{i} \bigg]$

$\dashrightarrow \bf \: \vec{S} = \frac{ { \lambda}^{2}v }{2\pi \epsilon_0 a}( \hat{j} \times \hat{i})$

$\dashrightarrow \bf \: \vec{S} = - \frac{ { \lambda}^{2}v }{2\pi \epsilon_0 a}( \hat{k})$

$\underline{\boxed{\bf {\purple{\: \vec{S} = - \frac{ { \lambda}^{2}v }{2\pi \epsilon_0 a}( \hat{k})}}}}$

Option (a) is the correct answer !!!