Field, Force, Wire, Charge - Performing This Thursday at 12:25 PM

Hmm...tomorrow is Wednesday, isn't it?

I think it's time for an AP question...think Thermo.


1.
a) 5 x 10^-5T
b) West
c) Net B-field is of magnitude 7.07 x 10^-5T directed at 45 degrees North of West. (It is not enough to say 7.07 x 10^-5T at 45 degrees!)

2. a = 0.0076 m/s^2

3. Since the force is attractive between the two wires, the current must be going in the same direction. k'I1I2/d = 6 x 10^-4 N/m...solving for the unknown current gives I = 48 A



4.
a) Remember that force/unit length means that from F = IBL, force per length is F/L = IB. Thus F/L = 2.5 A * 1.10 T * sin(90) = 2.75 N/m.

b) F/L = 2.5 A * 1.10 T * sin(45) = 1.944 N/m

5. F = IBL so I = F/(BL) = 11.67 A

6. Let's define a magnetic field OUT of the page to be positive.

a) B_total = kI1/d = (2 x 10^-7 T*m/A*10 A/.05 m = 4 x 10^-5T

b) B_total = k'I1/d + k'I2/d = (2 x 10^-7 T*m/A*10 A/.05 m + (2 x 10^-7 T*m/A*5 A/.05 m= 6 x 10^-5T

c) B_total = k'I1/d - k'I2/d = (2 x 10^-7 T*m/A*10 A/.05 m - (2 x 10^-7 T*m/A*5 A/.05 m= 2 x 10^-5T

7. F = IBL sin(60) = 1.212 N

8. This is an AP problem - don't be scared off by it! Remember to set a direction (in or out of the page) as positive and then sum all fields together, being sure to set fields positive or negative depending on the direction of each. We will use out of the page as positive.

Sum of fields at P:
-k'(1A)/(0.5 m) + k'(3A)/(1.0 m) = 2 x 10^-7 T

This shows that the field is directed out of the page (a) at the given magnitude of 2 x 10^-7 T.

c) Force is to the left, B is out of the page, and v is towards the top of the page. This only matches the left hand rule, which means the charge must be negative.

d) Fmag = qvB sin(90) so q = Fmag/(vB) = 5 x 10^-7 C

e) An electric force to the right of magnitude 10^-7 N would make the net force zero. Since Felec = qE, the field would have magnitude 0.2 N/C. Since the charge is negative, the electric field would need to be directed to the left for the force to be to the right.