Two long wires, 4cm apart, are parallel to each other. The wires have an equal a

Question

Two long wires, 4cm apart, are parallel to each other. The wires have an equal amount of current I1=I2=  6A flowing through them. Points P and R are a distance a = 3 cm above the top wire and below the bottom wire respectively. Point Q is halfway between the two wires. (Figure 1)

Part A

Part complete

What is the net magnetic field strength at point P in the figure? (Figure 1)

2.29x10^-5

Part B

Part complete

What is the net magnetic field's direction at point P in the figure?

out of the page

Part C

What is the net magnetic field strength at point Q in the figure? (Figure 1)

Part D

Part complete

What is the net magnetic field's direction at point Q in the figure?

into the page

Part E

What is the net magnetic field strength at point R in the figure?

Express your answer using three significant figures.

Part F

Part complete

What is the net magnetic field's direction at point R in the figure?

out of the page

Two long wires, 4cm apart, are parallel to each other. The wires have an equal amount of current I1=I2=  6A flowing through them. Points P and R are a distance a = 3 cm above the top wire and below the bottom wire respectively. Point Q is halfway between the two wires. (Figure 1)

Part A

Part complete

What is the net magnetic field strength at point P in the figure? (Figure 1)

2.29x10^-5

Part B

Part complete

What is the net magnetic field's direction at point P in the figure?

out of the page

Part C

What is the net magnetic field strength at point Q in the figure? (Figure 1)

Part D

Part complete

What is the net magnetic field's direction at point Q in the figure?

into the page

Part E

What is the net magnetic field strength at point R in the figure?

Express your answer using three significant figures.

Part F

Part complete

What is the net magnetic field's direction at point R in the figure?

out of the page

Explanation / Answer

Given

two parallel current carrying wires in opposite direcntions with I1=I2 = 6 A

separated by a distance of r =4 cm, a = 3 cm

Part A

point p is at a distance of 3 cm from I1 and 7 cm from I2

we know that the magnetic field due to a current carrying wire I at a distance r is  

B = mue0*I/2pi*r

B1 = mue0*I1/2pi*a , B2 = mue0*I2/2pi*(a+r)

B1 = 4pi*10^-7*6/(2pi*0.03) = 4*10^-5 T , directed out of the page By right hand rule

B2 = 4pi*10^-7*6/(2pi*0.07) T = 1.7143*10^-5 T, directed into the page by right hand rule

net field is B B1+B2 = 4*10^-5 -1.7143*10^-5 T = 2.2857*10^-5 T

PartB

the direction of the electric field is out of the page

PartC

magnetic field at Q is B1 = mue0*I1/2pi*r , B2 = mue0*I2/2pi*(r)

B1 = 4pi*10^-7*6/(2pi*0.02) = 6*10^-5 T , directed out of the page By right hand rule

B2 = 4pi*10^-7*6/(2pi*0.02) T = 6*10^-5 T , directed into the page by right hand rule

net field is B B1+B2 = 6*10^-5 T -6*10^-5 T = 0 T

PartD

the field direction is not defined

Part E

field at R

point R is at a distance of 3 cm from I2 and 7 cm from I1

we know that the magnetic field due to a current carrying wire I at a distance r is  

B = mue0*I/2pi*r

B2 = mue0*I2/2pi*a ,

B2 = 4pi*10^-7*6/(2pi*0.03) = 4*10^-5 T , directed out of the page By right hand rule

B1 = 4pi*10^-7*6/(2pi*0.07) T = 1.7143*10^-5 T, directed into the page by right hand rule

net field is B = B1+B2 = 4*10^-5 -1.7143*10^-5 T = 2.2857*10^-5 T

Part F

the direction is out of the page

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