Currents -I-1- and -I-2- flow in the wires shown in figure- The field is zero distance x to the right of O- Thenx-left- cfrac - - I - 1 - - - I - 2 - - right- ax-left- cfrac - - I - 2 - - - I - 1 - - right- ax-left- cfrac - - I - 1 - I - 2 - - - - I - 1 -I - 2 - - right- ax-left- cfrac - - I - 1 - I - 2 - - - - I - 1 -I - 2 - - right- a

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Toppr Admin · Accepted Answer

Magnetic field with both wire will be equal and opposite in diretion-x3BC-0I14-x3C0-a-x-x3BC-0I24-x3C0-a-x2212-x-a-x2212-xa-x-I2I1

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

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

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

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

Currents I1 and I2 flow in the wires shown in figure. The field is zero at distance x to the right of O. Thenx=(I1I2)ax=(I2I1)ax=(I1−I2I1+I2)ax=(I1+I2I1−I2)a

Magnetic field with both wire will be equal and opposite in diretion,μ0I14π(a+x)=μ0I24π(a−x)a−xa+x=I2I1

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

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

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

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