Question

Two long, straight wires are parallel and are separated by a distance of d = 0.210 m. The top wire in the sketch carries current I1 = 4.00 A , toward the right, and the bottom wire carries current I2 = 5.90 A , also to the right. At point P, midway between the two wires, what are the magnitude and direction of the net magnetic field produced by the two wires?

Answer 1

Answer:

$1.88\cdot 10^{-5} T$, inside the plane

Explanation:

We need to calculate the magnitude and direction of the magnetic field produced by each wire first, using the formula

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

where

$\mu_0$ is the vacuum permeability

I is the current

r is the distance from the wire

For the top wire,

I = 4.00 A

r = d/2 = 0.105 m (since we are evaluating the field half-way between the two wires)

so

$B_1 = \frac{(4\pi\cdot 10^{-7})(4.00)}{2\pi(0.105)}=7.6\cdot 10^{-6}T$

And using the right-hand rule (thumb in the same direction as the current (to the right), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is inside the plane

For the bottom wire,

I = 5.90 A

r = 0.105 m

so

$B_2 = \frac{(4\pi\cdot 10^{-7})(5.90)}{2\pi(0.105)}=1.12\cdot 10^{-5}T$

And using the right-hand rule (thumb in the same direction as the current (to the left), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is also inside the plane

So both field add together at point P, and the magnitude of the resultant field is:

$B=B_1+B_2 = 7.6\cdot 10^{-6} T+1.12\cdot 10^{-5}T=1.88\cdot 10^{-5} T$

And the direction is inside the plane.