Question

The value of magnetic field at origin due to current carrying wire ABC is : (coordinates of B is $\sqrt { 2 }$, O)

• A. $\frac { \mu _ { 0 } \mathrm { I } } { 8 \pi }$)
• B. $\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi } ( \sqrt { 2 } - 1 ) \hat { \mathrm { k } }$
• C.
• D.

Answer 1

Answer: (B)

$\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi } ( \sqrt { 2 } - 1 ) \hat { \mathrm { k } }$

Answer 2

B_AB = $\frac { \mu _ { 0 } \mathrm { I } } { 4 \pi ( \mathrm {~d} ) }$ (sina₁ – sina₂)

Where d is the ^ distance and a₁ and a₂ are the angles between ^ and ends of wire.

B_AB = [sin90⁰ – sin 45⁰]

= $\left[ 1 - \frac { 1 } { \sqrt { 2 } } \right]$ = $\frac { \mu _ { 0 } \mathrm { I } ( \sqrt { 2 } - 1 ) \hat { \mathrm { k } } } { 8 \pi }$

Magnetic field due to BC will be same and in same direction.

\ B₀ = $\frac { 2 \times \mu _ { 0 } \mathrm { I } ( \sqrt { 2 } - 1 ) } { 8 \pi }$