Question

The current in a solenoid with 23 turns per centimeter is 0.45A . The solenoid has a radius of 1.6cm . A long, straight wire runs along the axis of the solenoid, carrying a current of 15A . Part A Find the magnitude of the net magnetic field a radial distance of 0.80cm from the straight wire.

Answer 1

B=uNI/L
B=UNI/2r
B=UI/2pir

U=4pi x10^-7

Fm=qvBsin

Answer 2

What will be the field contributed by the straight wire alone?

Answer 3

what do you mean?

Answer 4

Well, the current through the wire is going to generate a magnetic field, isn't it? The net magnetic field will be the linear combination of the field due to the solenoid and the field due to the wire.

Answer 5

yes

Answer 6

So, do you know how to find the field contributed by the straight wire? That should be the easier of the two parts...

Answer 7

really sad answer, but no.

Answer 8

Damn, now I have to figure it out :-)

Answer 9

and another equation is Fm=LIBsin

Answer 10

for that I think the formula is
\[B = \frac{\mu_0 I}{2\pi r}\]

Answer 11

I was going to use that, but I thought it would only be is it was circular because of the 2pi r

Answer 12

i got 1.8 x10^-8

Answer 13

Well, the magnetic field lines make circles around the wire. If you are using your right hand and pointing your thumb along the wire in the direction of the current, the field lines follow your fingers, going in the direction they point.

Answer 14

oh yea, sorry every time it says straight conductor I think the field lines are going straight, bad mistake, sorry.

Answer 15

I got i got 1.8 x10^-8

Answer 16

Hmm, I got 0.000375 for my answer

Answer 17

what did you use for r and I?

Answer 18

\[B = \frac{4\pi * 10^{-7}*15}{2\pi *0.8*0.01} = \]

Answer 19

I = 15 A, r = 0.8 cm * 1 m/100 cm= 0.008 m

Answer 20

ok, I think your right. I made a mistake in my calculations

Answer 21

go on

Answer 22

so now we need to find the field inside the solenoid due to the solenoid (r = 1.6 cm > 0.8 cm)

Answer 23

ok, so B=UI/2pir equation

Answer 24

for that, the formula appears to be

\[B = \mu_0(\frac{N}{L})I\]

Answer 25

or that . I really don't like magnetism.

Answer 26

here I = 0.45 A and N/L = 23/1 cm or 23 / 0.01m

Answer 27

yeah, I'm not crazy about it either :-)

Answer 28

8.1x10^-8 ??????

Answer 29

but these field lines are running parallel to the straight wire. I'm not really sure how to combine them...

Answer 30

can you show me how you got that number?

Answer 31

{(4pix10^-7)(23 turns)/ (1.6)(100)}(0.45)

Answer 32

why are you including the radius? the field due to the solenoid is uniform within the core of the solenoid...

Answer 33

do i put 0.80 cm instead ?

Answer 34

http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/ampereslaw/solenoid.html

no, there's no radial component to worry about, just\[ B = \mu_0(\frac{N}{L})I\]

Answer 35

but what would L be?

Answer 36

N/L is turns per length. 23 turns in 1 cm or 23 turns in 0.01 m

Answer 37

I have to leave now, sorry I can't be of more help!

Answer 38

ok. Could you tell me what to do after I get the magnetic fields

Answer 39

I got 1.30 x10^-3