Question

. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer 1

Answer:

magnitude of net magnetic field at given point is

$B = 5 \times 10^{-6} T$

Explanation:

As we know that magnetic field due to a long current carrying wire is given as

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

here we we will find the magnetic field due to wire which is along x axis is given as

$i = 30 A$

r = 2 m

now we have

$B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}$

$B_1 = 3\times 10^{-6} T$ into the plane

Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

$i = 40 A$

r = 2 m

now we have

$B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}$

$B_2 = 4\times 10^{-6} T$ along + x direction

Since the two magnetic field is perpendicular to each other

So here net magnetic field is given as

$B = \sqrt{B_1^2 + B_2^2}$

$B = 5 \times 10^{-6} T$