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EMF and EMF Devices

Currents I1...

Question

Currents $I_{1}$ and $I_{2}$ flow in the wires shown in figure. The field is zero at distance $x$ to the right of $O$ . Then

A

x = (\frac{I_{1}}{I_{2}}) a

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B

x = (\frac{I_{2}}{I_{1}}) a

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C

x = (\frac{I_{1} - I_{2}}{{I_{1} + I}_{2}}) a

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D

x = (\frac{I_{1} + I_{2}}{{I_{1} - I}_{2}}) a

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Solution

The correct option is C $x = (\frac{I_{1} - I_{2}}{{I_{1} + I}_{2}}) a$
Magnetic field with both wire will be equal and opposite in diretion,
$\frac{μ_{0} I_{1}}{4 π (a + x)} = \frac{μ_{0} I_{2}}{4 π (a - x)}$

$\frac{a - x}{a + x} = \frac{I_{2}}{I_{1}}$
$\Rightarrow x = (\frac{I_{1} - I_{2}}{I_{1} + I_{2}}) a$

Hence Option C

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