619unMMtRH62_1210016709_jwk572

# 619unMMtRH62_1210016709_jwk572 - homework 08 KIM JI Due...

This preview shows pages 1–3. Sign up to view the full content.

homework 08 – KIM, JI – Due: Mar 16 2008, 4:00 am 1 Question 1, chap 29, sect 1. part 1 of 1 10 points Three very long wires are strung parallel to each other as shown in the figure below. Each wire is at the perpendicular distance 44 cm from the other two, and each wire carries a current of magnitude 6 . 5 A in the directions shown in the figure. I I I 3 2 1 z x y 44 cm 44 cm 44 cm × 3 2 1 y x z Cross-sectional View The permeability of free space is 1 . 2566 × 10 6 T · m / A. Find the magnitude of the net force per unit length exerted on the upper wire (wire 3) by the other two wires. Correct answer: 3 . 32623 × 10 5 N / m (toler- ance ± 1 %). Explanation: Let : r = 44 cm = 0 . 44 m and I = 6 . 5 A . Magnetic field due to a long straight wire is B = μ 0 I 2 π r , and the force per unit length between two parallel wires is F = μ 0 I 1 I 2 2 π r . There are two ways to solve this problem which are essentially the same. The first way is to find the net magnetic field at the upper wire from the two wires below ( vector B net = vector B 1 + vector B 2 ) and then find the force from vector F = I vector L × vector B . The crucial step here will be to add the magnetic fields as vectors . The second way would be to use vector F = I vector L × vector B to find the net force on the upper wire from the two lower wires vector F net = vector F 1 + vector F 2 , where we must be sure to add the forces as vectors. You should recognize that the two methods are formally identical. Let’s do it the first way. The magnitude magnetic field from wire 1 is found from Ampere’s law to be B 1 = μ 0 I 2 π r . Using the right hand rule the direction points up and to the left of wire as shown in figure 2. 30 30 60 60 60 B 1 B B 2 × 3 2 1 ˆ ˆ ı ˆ k Adding Magnetic Fields 30 30 60 60 60 F 13 F F 23 × 3 2 1 ˆ ˆ ı ˆ k Adding Forces Its components will then be vector B 1 = B [sin (30 cos (30 ı ] . Similarly, vector B 2 points down and to the left of wire 3; its components are given by vector B 2 = B [ sin (30 cos (30 ı ] .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
homework 08 – KIM, JI – Due: Mar 16 2008, 4:00 am 2 Notice that by symmetry the ˆ ( y ) compo- nent of the magnetic field vanishes. The net magnetic field is therefore vector B net = 2 μ 0 I cos (30 ) 2 π r ˆ ı . The force is then vector F = I vector L × vector B = I L parenleftbigg 2 μ 0 I cos (30 ) 2 π r parenrightbigg ( ˆ k × ˆ ı ) = I L parenleftbigg 2 μ 0 I cos (30 ) 2 π r parenrightbigg ( ˆ ) bardbl vector F bardbl bardbl vector L bardbl = μ 0 I 2 cos(30 ) π r = (1 . 2566 × 10 6 T · m / A) (6 . 5 A) 2 cos(30 ) π (0 . 44 m) = 3 . 32623 × 10 5 N / m . Notice: The direction of vector L is in the ˆ k direc- tion, since that is the direction of the current. After all the negative signs have cancelled, we notice that the force is in the ˆ direction.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern