Question

Two parallel, long wires carry currents $i _ { 1 }$ and $i _ { 1 } > i _ { 2 }$. When the currents are in the same direction, the magnetic field at a point midway between the wires is 10 μT. If the direction of $i _ { 2 }$ is reversed, the field becomes 30 μT. The ratio $i _ { 1 } / i _ { 2 }$ is

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

2

Answer 2

Initially when wires carry currents in the same direction as shown. Magnetic field at mid point O due to wires 1 and 2 are respectively

$B _ { 2 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 i _ { 2 } } { x }$◉

Hence net magnetic field at O

⇒ $10 \times 10 ^ { - 6 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 } { x } \left( i _ { 1 } - i _ { 2 } \right)$ .....(i)

If the direction of i₂ is reversed then

$B _ { 2 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 i _ { 2 } } { x } \otimes$

So $B _ { n e t } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 } { x } \left( i _ { 1 } + i _ { 2 } \right)$

⇒ $30 \times 10 ^ { - 6 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 } { x } \left( i _ { 1 } + i _ { 2 } \right)$ ......(ii)

Dividing equation (ii) by (i) $\frac { i _ { 1 } } { i _ { 2 } } = \frac { 2 } { 1 }$