Question

Current i flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at a distance r from O is

• A.
• B. $\frac { 2 \mu _ { 0 } \mathrm { i } } { \pi \mathrm { r } }$
• C. $\frac { \mu _ { 0 } \mathrm { i } } { 4 \pi \mathrm { r } } ( \sqrt { 2 } + 1 )$
• D. $\frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \mathrm { i } } { \mathrm { r } } ( \sqrt { 2 } + 1 )$

Answer 1

Answer: (D)

$\frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \mathrm { i } } { \mathrm { r } } ( \sqrt { 2 } + 1 )$

Answer 2

By using $B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r } \left( \sin \phi _ { 1 } + \sin \phi _ { 2 } \right)$, from figure

$d = r \sin 45 ^ { \circ } = \frac { r } { \sqrt { 2 } }$

Magnetic field due to each wire at P

$B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { ( r / \sqrt { 2 } ) } \left( \sin 45 ^ { \circ } + \sin 90 ^ { \circ } \right)$

$= \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r } ( \sqrt { 2 } + 1 )$

Hence net magnetic field at P