Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire?

Respuesta :

Answer:

[tex]B_{net} = 3.3 \times 10^{-5} T[/tex]

Explanation:

Magnetic field due to straight long wire is always perpendicular to the line joining the wire and the point

it is given by the equation

[tex]B = \frac{\mu_0 i}{2\pi r}[/tex]

now the magnetic field due to 30 A current is given as

[tex]B_1 = \frac{(2\times 10^{-7})(30)}{0.15}[/tex]

[tex]B_1 = 4\times 10^{-5} T[/tex]

Now magnetic field due to other wire having current 40 A is given as

[tex]B_2 = \frac{(2 \times 10^{-7})(40)}{0.25}[/tex]

[tex]B_2 = 3.2 \times 10^{-5} T[/tex]

now net field along Y direction is given as

[tex]B_y = B_2cos37[/tex]

[tex]B_y = 2.56 \times 10^{-5} T[/tex]

now for X direction magnetic field we know

[tex]B_x = B_1 - B_2sin37[/tex]

[tex]B_x = (4\times 10^{-5}) - (3.2 \times 10^{-5}sin37)[/tex]

[tex]B_x = 2.07 \times 10^{-5} T[/tex]

Now net magnetic field is given as

[tex]B_{net} = \sqrt{B_x^2 + B_y^2}[/tex]

[tex]B_{net} = \sqrt{2.56^2 + 2.07^2} \times 10^{-5} T[/tex]

[tex]B_{net} = 3.3 \times 10^{-5} T[/tex]

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