PHY2054 Section 4350
Quiz 6: Ch 19
October 20, 2011
NAME
UFID
1. The cube is 50.0 cm on each edge. Four straight segments of wire - ab, bc, cd, and da - form a closed loop that
carries a current I = 5.00 A in the direction shown. A uniform magnetic field of magnitude B = 0.020 T is in
the positive y-direction.
(a) Determine the magnitude and direction of the magnetic force on each segment. (3.5/5)
(b) What’s the total magnetic force on the loop? (0.5/5)
Answer:
(a)
i.
Fab = 0,
since ab and B are anti-parallel.
ii.
Fbc = ILB = 5 × 0.5 × 0.02 = 0.05(N ),
along the negative x-direction, i. e. −x̂.
iii.
√
Fcd = I ( 2L) B sin (45◦ ) = ILB = 0.05(N ),
along the negative z-direction, i. e. −ẑ.
iv.
√
Fda = I ( 2L) B ≈ 0.07(N ),
√
along the direction (x̂ + ẑ)/ 2.
Therefore,
√
x̂ + ẑ
Fda = I ( 2L) B √ = 0.05(x̂ + ẑ)(N ).
2
2
(b)
F = Fab + Fbc + Fcd + Fda
= 0 − 0.05x̂ − 0.05ẑ + 0.05(x̂ + ẑ)
= 0.
Actually, without calculation, we know F = 0 because the magnetic force on a current loop in a uniform
magnetic field must be zero.
2. When a magnetic field causes a charged particle to move, which of following is true?
(a) The energy of the particle changes.
(b) The momentum of the particle changes.
(c) Both energy and momentum of the particle change.
(d) Neither energy nor momentum of the particle changes.(0.5/5)
3. An electron is moving north in a region where the magnetic field is north. The magnetic force exterted on the
electron is
(a) zero
(b) up
(c) down
(d) east
(e) west. (0.5/5)
formula:
force: F = qv × B;F = Il × B.
0I
magnetic field: B = µ2πr
(infinitely long straight line); B =
r = mv/qB; g = 9.8m/s2 .
µ0 I
2r
(circular loop).