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>>Moving Charges and Magnetism
>>Problems on Biot-Savart Law
>> $Currents I1 and I2 flow in the wires sho$

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

166819

A

$x = (\frac{I _{1}}{I _{2}}) a$

B

$x = (\frac{I _{2}}{I _{1}}) a$

C

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

D

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

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Solution

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Correct option is C)

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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