Question

The drawing shows two long, thin wires that carry currents in the positive z direction. Both wires are parallel to the z axis. The 50-A wire is in the x-z plane and is 5 m from the z axis. The 40-A wire is in the y-z plane and is 4 m from the z axis. What is the magnitude of the magnetic field at the origin?

Answer 1

Answer:

The magnitude of the magnetic field at the origin is $2.56\times 10^{-6}\ T$.

Explanation:

Given :

50-A wire is in the x-z plane and is 5 m from the z axis.

Also , 40-A wire is in the y-z plane and is 4 m from the z axis.

Now , since both the wire are perpendicular to each other .

Therefore , magnetic field are also perpendicular to each other .

Magnetic field at origin due to wire 1 is :

$B_1=\dfrac{\mu_oI_1}{2\pi R_1}\\\\B_1=\dfrac{(50)\mu_o}{2\pi( 5)}\\\\B_1=\dfrac{5\mu_o}{\pi}$

Magnetic field at origin due to wire 2 is :

$B_2=\dfrac{\mu_oI_2}{2\pi R_2}\\\\B_2=\dfrac{(40)\mu_o}{2\pi( 4)}\\\\B_2=\dfrac{4\mu_o}{\pi}$

Now , therefore net magnetic field is :

$B=\sqrt{B_1^2+B_2^2}\\\\B=\sqrt{(\dfrac{5\mu_o}{\pi})^2+(\dfrac{4\mu_o}{\pi})^2}\\\\B=\dfrac{\sqrt{41}\mu_o}{\pi}$

Putting value of $\mu_o=4\pi \times 10^{-7}\ H/m$

We get ,

$B=\sqrt{41}\times 4\times 10^{-7}\\B=2.56\times 10^{-6}\ T$

Therefore, the magnitude of the magnetic field at the origin is $2.56\times 10^{-6}\ T$.