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Phys 1120, CAPA #10 solutions.
1) This problem is about B fields from wires BOTH wires produce Bfields at ALL points in
space. So when you look at, say, point p1, it receives contributions from the upper and lower
wire. In this case (since the currents go in opposite directions, those fields will also go in
opposite directions, that's how it's possible to get the field to be 0 down there at point p1!)
The point p1 is "a" from the lower wire, and "3a" from the upper wire. Using the right hand rule,
the B field due to wire I1 will be INTO the page down there where p1 is. The B field due to wire
I2 will be OUT of the page down there where p1 is. (Convince yourself, don't just take my word
for it!! If you don't know how to find the direction of B, please ask someone, it's hard to write
but easy to show with fingers! see Fig 32.14 to remind yourself of the basic picture. )
So, the fields can cancel. Formulawise
B1 = mu0 I1 / (2 pi (3a)) IN.
B2 = mu0 I2/ (2 pi (a)) OUT.
In order for the field to vanish, those two must be equal in magnitude, so
I1/3a = I2/a, or I1 = 3 I2. They gave us I2, so just triple it to find I1.
(Makes sense, I1 is three times farther away, so the current needs to be 3 times bigger to give an
equal field, which is what you need to get cancellation!)
11) In the middle, at point P2, the B field due to I1 is still into the page, but the B field due to I2
is now ALSO into the page. (We're now ABOVE wire 2). Again, convince yourself! In the "in
between" region, the two B fields add, they don't cancel. So we have
B1 + B2 = muo (I1 + I2)/(2 pi a) (where, in this formula, I really mean the magnitude of I1 and
the magnitude of I2, I've already accounted for their directions when I worked out that they add
up) Since I1 = 3I2, we have mu0*4I2/(2 pi a) ( Just plug in for CAPA)
3) Magnetic fields form circular loops around currents, in the "right hand sense". (See e.g. Fig
32.14 in your text). In this case that means B at point p1 points SOUTH (down the page), and the
same at p2 (but, weaker, because it's farther away) An electron at REST will feel no magnetic
force (you have to be moving to feel a force!)
4) Since B = mu0*I/(2 pi r), if you are twice as far away, the B field is half as strong. (It varies
inversely as radius, not radius squared or anything) So, depending on which point was given to
you (p1 or p2) just divide or multiply the number given by two!
5) I = dq/dt. Here, dq = 15 C (your number might vary), and dt = 1.5E3 sec, so I = 15/1.5E3
A = 10^4 Amps. (Big current!) Now just use the formula 305 of your text (same as 308),
B = mu0*I/ (2 pi r). We just got I, mu0 is a constant, r is given. The answer comes out pretty
small  magnetic fields TEND to be small in nature, unless something rather unusual is going on
(it's hard to get "1 Tesla" fields!)
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View Full Document 6) A) The long wire generates a B field around itself (RHR #1) which circles around the long
wire, and thus points OUT of the plane of the paper everywhere in the picture, in particular,
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This note was uploaded on 04/28/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.
 Fall '08
 ROGERS

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