Question

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?

Answer 1

Answer:

$B_{net} = 3.3 \times 10^{-5} T$

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

$B = \frac{\mu_0 i}{2\pi r}$

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

$B_1 = \frac{(2\times 10^{-7})(30)}{0.15}$

$B_1 = 4\times 10^{-5} T$

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

$B_2 = \frac{(2 \times 10^{-7})(40)}{0.25}$

$B_2 = 3.2 \times 10^{-5} T$

now net field along Y direction is given as

$B_y = B_2cos37$

$B_y = 2.56 \times 10^{-5} T$

now for X direction magnetic field we know

$B_x = B_1 - B_2sin37$

$B_x = (4\times 10^{-5}) - (3.2 \times 10^{-5}sin37)$

$B_x = 2.07 \times 10^{-5} T$

Now net magnetic field is given as

$B_{net} = \sqrt{B_x^2 + B_y^2}$

$B_{net} = \sqrt{2.56^2 + 2.07^2} \times 10^{-5} T$

$B_{net} = 3.3 \times 10^{-5} T$