COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 19, Problem 70QAP
To determine

(a)

The magnitude of magnetic field at y =0 cm according to the adapted coordinate system

Expert Solution
Check Mark

Answer to Problem 70QAP

The magnitude of the magnetic field at (y =0 cm) = 0 T

Or in other words there is no magnetic field at y= 0 cm

Explanation of Solution

Given:

Two long straight wires are positioned such that they are parallel to x at y=2.5 cm and y= -2.5 cm as shown in the diagram below. Each wire carries a current of 16 A.

Formula used:

  B=μ0i2πr(a)where B= Magnitude of the magentic field due to a long, straight wireμ0= permeability of free spacer = distance from the wire to the location where the field is measured

Right hand rule for the field directionIf you point your right thumb in the direction of the current and curl your fingers, the magnetic field curls around the field lines in the direction of the curled fingers of your right hand.

Calculation:

One can make use of the superposition to calculate the magnitude of the magnetic field at designated location of the y axis.

Let us denote the magnitude of the magnetic field generated by the top wire as Btopand the magnetic field generated by the wire below as Bbottom.

Applying the right hand rule one could clearly see that Btop acts in the -z direction while Bbottomacts in the +z direction.

Let B= total magnitude of the magnetic field at y=0 cm

Then B = Btop + Bbottom

Substituting for equation (a) we could find B as follows where rtop= distance from the top wire to y=0 and rbottom= distance from the bottom wire to y=0.

  B = ( μ 0i2π r top) + ( μ 0i2π r bottom)B =( μ 0i2π){1r top + 1r bottom}rtop=rbottom=2.5 cm = 0.025m ;B=( μ 0i2π){10.025 m + 10.025 m}B=0T

Conclusion:

The magnitude of the magnetic field at (y =0 cm) = 0 T

Or in other words there is no magnetic field at y= 0 cm

To determine

(b)

The magnitude of magnetic field at y = 1.0 cm according to the adapted coordinate system

Expert Solution
Check Mark

Answer to Problem 70QAP

The magnitude of the magnetic field at (y =1.0 cm) = (-1.2 x 10-4)T

Explanation of Solution

Given:

Two long straight wires are positioned such that they are parallel to x at y=2.5 cm and y= -2.5 cm as shown in the diagram below. Each wire carries a current of 16 A.

Calculation:

One can make use of the superposition to calculate the magnitude of the magnetic field at designated location of the y axis.

Let us denote the magnitude of the magnetic field generated by the top wire as Btopand the magnetic field generated by the wire below as Bbottom.

Applying the right handrule, one could clearly see that Btop acts in the -z direction while Bbottomacts in the +z direction.

Let B= total magnitude of the magnetic field at y=1.0 cm

Then B = Btop + Bbottom

Substituting for equation (a) we could find B as follows where rtop= distance from the topwire to y=1.0cm and rbottom= distance from the bottom wire to y=1.0 cm.

  B = ( μ 0i2π r top) + ( μ 0i2π r bottom)B =( μ 0i2π){1r top + 1r bottom}rtop=1.5 cm = 0.015 m  and rbottom=3.5 cm = 0.035 mB=( μ 0i2π){10.015 m + 10.035 m}B =((4π× 10 7 T.m A)×(16A)2π){10.015 m + 10.035 m}B = (-1.2 ×104) T 

Conclusion:

The magnitude of the magnetic field at (y =1.0 cm) = (-1.2 x 10-4)T

To determine

(c)

The magnitude of magnetic field at y = 4.0 cm according to the adapted coordinate system

Expert Solution
Check Mark

Answer to Problem 70QAP

The magnitude of the magnetic field at (y =4.0 cm) = (2.6 x 10-4)T

Explanation of Solution

Given:

Two long straight wires are positioned such that they are parallel to x at y=2.5 cm and y= -2.5 cm as shown in the diagram below. Each wire carries a current of 16 A.

Calculation:

One can make use of the superposition to calculate the magnitude of the magnetic field at designated location of the y axis.

Let us denote the magnitude of the magnetic field generated by the top wire as Btopand the magnetic field generated by the wire below as Bbottom.

Applying the right hand rule, one could clearly see that Btop acts in the +z direction while Bbottomalso acts in the +z direction.

Let B= total magnitude of the magnetic field at y=4.0 cm

Then B = Btop + Bbottom

Substituting for equation (a) we could find B as follows where rtop= distance from the top wire to y=4.0cm and rbottom= distance from the bottom wire to y=4.0 cm.

  B = ( μ 0i2π r top) + ( μ 0i2π r bottom)B =( μ 0i2π){1r top + 1r bottom}rtop=1.5 cm = 0.015 m  and rbottom=6.5 cm = 0.065 mB=( μ 0i2π){10.015 m + 10.065 m}B =((4π× 10 7 T.m A)×(16A)2π){10.015 m + 10.065 m}B = (2.6 ×104) T 

Conclusion:

The magnitude of the magnetic field at (y =4.0 cm) = (2.6 x 10-4)T

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Chapter 19 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

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