Two long straight parallel conductors carry currents ' I 1 ′ and ' I 2 ′ ( I 1 I 2 ) . When direction of ' I 1 ' and ' I 2 ' is same, the magnetic field int

Question

Two long straight parallel conductors carry currents ' I 1 ​ ′ and ' I 2 ​ ′ ( I 1 ​ > I 2 ​ ) . When direction of ' I 1 ​ ' and ' I 2 ​ ' is same, the magnetic field intensity midway between two conductors is 20 μ T . If direction of ' I 2 ​ ' is reversed, the field intensity becomes 50 μ T . The ratio I 2 ​ / I 1 ​ is

TestHub · Accepted Answer

Let the distance between the two conductors be $2 r$ . The magnetic field at the midpoint due to a current $I$ is $B = \frac{μ _{0} I}{2 π r}$ .

When the currents are in the same direction, the magnetic fields at the midpoint are in opposite directions. Since $I_{1} > I_{2}$ , the net magnetic field is:

$B_{1} = \frac{μ _{0}}{2 π r} (I_{1} - I_{2}) = 20 μ T$

When the direction of $I_{2}$ is reversed, the currents are in opposite directions, so the magnetic fields at the midpoint are in the same direction. The net magnetic field is:

$B_{2} = \frac{μ _{0}}{2 π r} (I_{1} + I_{2}) = 50 μ T$

Dividing the two equations, we get:

$\frac{I _{1} - I _{2}}{I _{1} + I _{2}} = \frac{20}{50} = \frac{2}{5}$

$5 I_{1} - 5 I_{2} = 2 I_{1} + 2 I_{2}$

$3 I_{1} = 7 I_{2}$

$\frac{I _{2}}{I _{1}} = \frac{3}{7}$

Answer: $\frac{3}{7}$

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