Physics, 06.06.2020 14:57 hit89

Two wires carry current I1 = 47 A and I2 = 29 A in the opposite directions parallel to the x-axis at y1 = 9 cm and y2 = 15 cm. Where on the y-axis (in cm) is the magnetic field zero?

Answers: 3

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Answers

Approximately $25\; \rm cm$ , assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.

Explanation:

Note that y_1 < y_2 , which means that the first wire (with current I_1 ) is underneath the second wire (with current I_2 .) Let denote the -coordinate (in $\rm cm$ ) of a point of interest. The two wires partition this region into three parts:

The region under the first wire, where y < y_1

;The region between the first and the second wire, where y_1 < y < y_2

; andThe region above the second wire, where y y_2

Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that I_1 points to the left while I_2 points to the right. Let B_1 and B_2 denote the magnetic field due to I_1 and I_2 , respectively.

In the region below the first wire, B_1

points out of the x y

-plane while B_2

points into the x y

-plane.In the region between the two wires, both B_1

and

point into the x y

-plane.In the region above the second wire, B_1

points into the x y

-plane while B_2

points out of the x y

-plane.

The (net) magnetic field on this plane would be zero only in regions where B_1 and B_2 points in opposite directions. That rules out the region between the two wires.

At a distance of away from a wire with current and infinite length, the formula for the magnitude of the magnetic field due to that wire is:

$\displaystyle B = \frac{\mu\, I}{2\pi\, R}$ ,

where $\mu$ is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of $\mu$ does not affect the answer to this question, as long as it is constant throughout this region.

Note that the value of in this formula is supposed to be positive. Let R_1 and R_2 denote the distance between the point of interest and the two wires, respectively.

In the region under the first wire, R_1 = y_1 - y = 9 - y

, while

. In the region above the second wire, R_1 = y - 9

, while

Make sure that given the corresponding range of , these distances are all positive.

Strength of the magnetic field due to the first wire: $\displaystyle B_1 = \frac{\mu\, I_1}{4\pi\, R_1}$ .Strength of the magnetic field due to the second wire: $\displaystyle B_1 = \frac{\mu\, I_2}{4\pi\, R_2}$ .

For the net magnetic field to be zero at a certain -value, the strength of the two magnetic fields at that point should match. That is:

B_1 = B_2 .

$\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}$ .

Simplify this equation:

$\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}$ .

In the region under the first wire (where y < 9

,) this equation becomes $\displaystyle \frac{47}{9 - y} = \frac{29}{15 - y}$ .In the region above the second wire (where y 15

,) this equation becomes $\displaystyle \frac{47}{y - 9} = \frac{29}{y - 15}$ .

These two equations give the same result: y = 25 . However, based on the respective assumptions on the value of , this value corresponds to the region above the second wire.

Answer from: Quest

coil wire, increase current, put iron inside the coil

Answer from: Quest

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

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Two wires carry current I1 = 47 A and I2 = 29 A in the opposite directions parallel to the x-axis at...