Physics, 22.07.2019 18:10 Valfrutos2912

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?

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Answer from: lazavionadams81

The magnitude and direction of the net magnetic field generated by the two wires will be $1.55\times 10^{-8}$ .

What is magnetic field strength?

The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.

$B = \frac{u_00I}{2\pi r}$

(I) is the current

r is the distance from the probe

B is the induced magnetic field

r denotes the distance between the wire and the object.

u_0 is the permeability to vacuum

For magnetic field in the top wire

Given

0.105 m = r = d/2

$B_1= \frac{4\pi\times10^{-7}\times4.00}{2\pi \times 0.105}$

$\rm{B_1=7.6\times 10^{-6}}$ T

As the current to the left other fingers wrapped around the thumb showing the direction of the magnetic field lines.

Given data for bottom wire

I= 5.90 A

r = 0.105 m r = 0.105 m r = 0.105 m

$B_2= \frac{4\pi\times10^{-7}\times5.90}{2\pi \times 0.105}$

$\rm{B_2=1.12 \times 10^-5$ T

As the current to the left other fingers wrapped around the thumb showing the direction of the magnetic field lines. The field lines direction at point P is also inside the plane.

The megnitude of the resultant magnetic field be the sum of both the field

$\rm{B=B_1+B_2}$

$\rm{B=7.6\times 10^{-6}+ 1.12\times 10^{-5}}$

$B = 1.88\times 10^{-5}$ T

Hence The megnitude of the resultant magnetic field be the sum of both the fields $1.88\times 10^{-5}$ T.

To learn more about the strength of induced magnetic field refer to the link;

link

Answer from: suewignall

$1.88\cdot 10^{-5} T$ , 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)

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

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

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

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.

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