Can someone please check my answers for Physics due tonight?!?!?!? Please show y
ID: 1270214 • Letter: C
Question
Can someone please check my answers for Physics due tonight?!?!?!?
Please show your steps so I can compare with my work and see where I went wrong so I can learn :)
I know that you have to use the law of cosine with this problem. Please try your best and don't just rush through this problem :) I really appreciate your help
THE PHYSICS PROBLEM (WITH PICTURE ATTACHED)
Two long thin parallel wires13.0 cm apart carry 29-A currents in the same direction.
Part A
Determine the magnitude of the magnetic field vector at a point 10.0?cm from one wire and 6.0?cm from the other (Figure 1) .
Express your answer using two significant figures
B = .............T
Part B
Determine the direction of the magnetic field vector at that point.
Express your answer using two significant figures.
?= ....... ?, measured counterclockwise from the positive x axis
MY WORK
B1x=B0cos?1
B1y=B0sin?1
where
B0=?0I/(2?d1)
d1=6 cm =0.06 m
?1=39.7
B1X = 7.4375 X 10^-5
B1Y= 6.174755 X 10^-5
B2x=-B0cos?2
B2y=B0sin?2
where
B0=?0I/(2?d2)
d2=10 cm = 0.1 m
?2=62.6
B2X = 2.669 X 10^-5
B2Y = 5.1493 X 10^-5
Bx=Bx1+Bx2
By=By1+By2
BX= 1.01065 X 10^-4
BY= 1.1324 X 10^-4
B=(Bx^2+By^2)^1/2
B= 1.5178 X 10^-4 (Unit of T)
b)
direction is
?=arctan(By/Bx)
?=48.25
Update to work
Applying cosine rule to triangle APB. I'll guess ?1 is QAP
10
Explanation / Answer
For their magnitudes, we have
B1= (mo/4p)2I/r1
= (10-7T.m/A)2(29A)/(0.10m) = 5.8 x 10-5T.
B2 = (mo/4p)2IB/r2
= (10-7T.m/A)2(29A)/(0.060m) = 0.9667 x 10-4T.
We use the property of the triangle to find the angles shown:
r22 = r12 + d2 - 2r1dcosq1;
(6.0cm)2 = (10.0cm)2 + (13.0cm)2 - 2(10.0cm)(13.0cm)cosq1,
which gives cosq1 = 0.896, q1 = 26.36