PHY2323 Assignment 4 Solutions
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PHY2323
Assignment 4 Solutions
Question 1
To start we sketch a diagram of the situation. For a negative charge travelling in the x-direction with a B-field in the y-direction will experience a force in the negative z-direction.
O2
z
B
x
F
The force acting on a moving charged particle in a magnetic field (Lorentz Force) is given by the expression:
O1 = q ×
For a charge Q = 一6.4 × 10 — 19 C with velocity = v = 20 m s — 1 in a magnetic field = Bgˆ = 5 Tgˆ the force is:
O3
= QvB( × gˆ) = QvB之ˆ = 一6.4 × 10 — 17 N之ˆ
The centripetal acceleration of circular motion is a = v2 /R (which occurs whenever the force is acting perpendicular to the particles velocity), applying Newton’s 2nd law relates the force to this acceleration:
O1 |F | = m|a| = = |QvB|
rearranging allows the radius to be found.
O3 R = = = │ │ = 2.25 × 10 — 11 m
Question 2
z
(0,0,3)
y
(0,-1,0) (0,1,0)
There is no symmetry here that would allow us to use Ampere’s law and so we must use the Biot-Savart Law.
O1
=
Id × ( 一 r-′ )
| 一 r-′ |3
Each segment of the triangle can be evaluated separately and the field form each can then be added together at the end since magnetic fields are linear.
O2 Considering the segment between (0, -1, 0) and (0, 1, 0) we find that = 0, r-′ = ygˆ and d = dygˆ. This
means that d × ( 一 r-′ ) = 一ydygˆ × gˆ = 0 and so there is no magnetic field due to this segment at the origin.
O1
O1
O1 |
For the segment (0, 1, 0) to (0, 0, 3), = 0, r-′ = ygˆ + z之ˆ and d= dygˆ + dz之ˆ. This then gives: d × ( 一 r-′ ) = 一(dygˆ + dz之ˆ) × (ygˆ + z之ˆ) = (ydz 一 zdy) The constraint of the line is that z = 3 一 3y and so dz = 一3dy. This gives: d × ( 一 r-′ ) = (一3ydy 一 (3 一 3y)dy) = 一3dy and | 一 r-′ |3 = ╱y2 + z2、尸(g) = ╱y2 + (3 一 3y)2、尸(g) = ╱ 10y2 一 18y + 9、尸(g) = 10 ′10 ╱(y 一 0.9)2 + 0.09、尸(g) substituting this into the Biot-Savart law results in: 2 = µ0 0 一3Idy 4π 1 10 ′ 10 ((y 一 0.9)2 + 0.09) 尸(g) making the substitution u = y 一 0.9 2 = 一3µ0 I —0 à 9 |
|
2 =
40(一)π(3)′(µ0) ┌ ┐
O2
2 =
40(一)π(3)′(µ0) ╱ 一103′ 10 一 \ =
O1 |
From symmetry we can see that the line from (0, 0, 3) to (0,-1, 0) will contribute the same field. Hence the total field at the origin will be: |
O1
= = 1.33 × 10 —7 T
Question 3
(a) We use the Biot-Savart Law
O1
=
Id × ( 一 r-′ )
| 一 r-′ |3
For the partial circular section da we evaluate in cylindrical coordinates: = 0, r-′ = Rpˆ and d = Rdφ. This gives:
O1
d × ( 一 r-′ ) = 一R2 dφ × pˆ = R2 dφ之ˆ
applying to the Biot-Savart law gives:
= µ0 5甘|4 IR2 dφ之ˆ
4π —甘|4 R3
O1
O1
O1
O1
= 之ˆ
2022-04-01