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Question: An infintely long wire carrying current I is along y-axis such that its one end is at point A(0, b) ...

An infintely long wire carrying current I is along y-axis such that its one end is at point A(0, b) while the wire extends upto + . The magnitude of magnetic field strength at point (a, 0).

A

μ0I4πa(1+ba2+b2)\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi \mathrm { a } } \left( 1 + \frac { \mathrm { b } } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)

B

μ0I4πa(1ba2+b2)\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi \mathrm { a } } \left( 1 - \frac { \mathrm { b } } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)

C

μ0I4πa(ba2+b2)\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi \mathrm { a } } \left( \frac { \mathrm { b } } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)

D

None of these

Answer

μ0I4πa(1ba2+b2)\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi \mathrm { a } } \left( 1 - \frac { \mathrm { b } } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)

Explanation

Solution

B = μ04π\frac { \mu _ { 0 } } { 4 \pi } ia\frac { \mathbf { i } } { \mathrm { a } } (sin 90ŗ + sin(–q))

= μ04π\frac { \mu _ { 0 } } { 4 \pi } (1ba2+b2)\left( 1 - \frac { b } { \sqrt { a ^ { 2 } + b ^ { 2 } } } \right)