Magnetic Field Review
1.
Dr. Ray Kwok
Find the magnitude of the induced current I(t) on 20 turns of wire with internal resistance =
r r
5Ω if the time-varying magnetic field through the loops is
B (r , t ) = (2 xˆ − zˆ )sin( 40t ).
Use Lenz’s Law to figure the direction of the current. Is it positive or negative for small t
according to the direction drawn in figure?
y
I(t)
4cm
3cm
20 turns
x
r
A = −0.0012 zˆ
r r
Φ = B ⋅ A = 0.0012 sin( 40t )
dΦ
ε = −N
= −0.96 cos( 40t )
dt
I (t ) =
ε
R
= −0.192 cos( 40t )
A
The sign is “−”ve according to Lenz’s Law.
For small t, B is increasing in the –z direction, so induced B is in +z direction.
And induced current is counter-clockwise.
Magnetic Field Review
2.
Dr. Ray Kwok
A semi-circle of radius 2 cm with its center at the origin as shown is carrying a steady
current of 5 A. Calculate the magnitude AND direction of the magnetic field at (0,0,4)cm
in this coordinate.
y
R
x
r
z
α
dB
Bz = −
r
r µ o Idl × rˆ µo IRdθ sin(90 o )
dB =
=
4π r 2
4π
r2
µ o IRdθ R
µ o IR 2 dθ
dBz = dB sin α = −
=−
4π r 2 r
4π r 3
µ IRdθ z
dBxy = dB cos α = o
4π r 2 r
µ zIR sin θdθ
dBy = dBxy sin θ = − o
4π
r3
µ o IR 2
µo (5)(0.02)2
µo IR 2 π
d
θ
=
−
=
−
= −5.59 µo
4πr 3 ∫0
4( R 2 + z 2 )3 / 2
4(0.022 + 0.042 )3 / 2
µo zIR
µ (0.04)(5)(0.02)
µ o zIR π
By = −
sin θdθ = −
=− o
= −7.12µ o
3 ∫
2
2 3/ 2
4π r 0
2π ( R + z )
2π (0.02 2 + 0.04 2 ) 3 / 2
r
B = µ o (0,−7.12,−5.59) = 9.05µo ∠52 o from –z to –y.
Magnetic Field Review
Dr. Ray Kwok
Note: if you still have trouble seeing the angle, consider this…..
r
dl = Rdθ sin θ ,− cos θ ,0
r
r = − R cos θ ,− R sin θ , z
y
dl
R
θ
x
r = R 2 cos 2 θ + R 2 sin 2 θ + z 2 = R 2 + z 2
r
rˆ = r / r
r
r µ o Idl × rˆ µo IRdθ
dB =
=
4π r 2
4π r 3
xˆ
sin θ
− R cos θ
µo IRdθ
(− z cos θ )
4π r 3
µ IRdθ
dBy = o
(− z sin θ )
4π r 3
µ IRdθ
(− R )
dBz = o
4π r 3
µ o IRz π
Bx = −
dθ cos θ = 0
4π r 3 ∫0
dBx =
µ IRz
µ o IRz π
By = −
dθ sin θ = − o 3
3 ∫
2π r
4π r 0
µ o IR 2 π
µo IR 2
Bz = −
dθ = −
4π r 3 ∫0
4 r3
which is the same as before…
yˆ
− cos θ
− R sin θ
zˆ
0
z
Magnetic Field Review
3.
Dr. Ray Kwok
The triangular loop of wire shown in Figure carries a current I = 5 A in the direction shown.
The loop is in a uniform magnetic field that has magnitude B = 3 T along the y-axis as
shown.
a. Find the force exerted by the magnetic field on each side of the triangle (magnitude and
direction).
b. What is the net force on the loop?
c. The loop is pivoted about an axis along ab in the diagram. Calculate the net torque of
the loop (magnitude and direction).
d. What is the magnetic moment of the loop?
e. Calculate the torque about the axis ab by using µ x B.
f. Describe the direction of the rotation.
c
θ
B, y
L
x
3m
(a) Fba = (5)(4)(3) = 60 N (-z)
Fac = 0
Fcb = (5)L(3)sinθ = (5)(4)(3) = 60 N (+z)
(b) Net Force = 0
I
a
4m
b
(c) Net Torque = Fcb (3/2) = 90 N-m (+x, out
the page from top)
(d) µ = IA = (5)(3)(4)/2 = 30 Am2 (-z)
o
(e) Torque = µ x B = µBsin90 = 90 N-m (+x)
(f) The loop rotates about the x-axis.
(out the page above ab)
Magnetic Field Review
4.
Dr. Ray Kwok
Point P is at a distance R away from a straight wire of length L carrying a current I as
shown in figure. (a) What is the magnetic field vector at point P? (b) If the wire is then
extended to a semi-circle and connect to another straight wire of length L (shown in the
lower figure), what would the magnetic field at point P become?
r
r µ o Idl × rˆ µo Idx sin θ µ o IRdx
(a)
(a) dB =
=
=
⊕
I
x
4π r 2
4π
r2
4π r 3
θ
µ o IRdx µ o IR L dx
R
B=∫
=
4π r 3
4π ∫0 r 3
P
cot θ =
sin θ =
x
R
1 sin θ
=
r
R
− R csc 2 θdθ = dx
(b)
R
r
3
θ
dx
1 m
 sin θ 
2
=
−
R
csc
d
sin θdθ
=
−
θ
θ


∫0 r 3 ∫
R 2 π∫/ 2
 R 
L
I
R
P
(
)


1
[cosθ ]θπm/ 2 = 12  2 L 2 
2
R
R  L +R 
0
r µ IR 1 
 µo I 
L
L

=

B= o
2 


2
2
2
4π R  L + R  4πR  L + R 2
L
dx
∫r
3
=
(b) B due to a semi-circle (at the center) = ½ (µoI/2R) = µoI/4R.
r
µ I 
 µ o I
L
 +
Btotal = 2  o 
 4πR  L2 + R 2  4 R
r

2L
µI
Btotal = o 1 +
⊕
4 R  π L2 + R 2 

⊕


Magnetic Field Review
5.
Dr. Ray Kwok
A long straight copper wire of gauge (AWG) # 15 (diameter = 1.45 mm) is carrying a
uniform current density given by J = 8000 (A/m2). The resistivity of copper is 1.72 x 10-6
Ω-cm.
a. What is the magnetic field (vector) created by this current at 0.5 m away from the
center of the wire?
b. If an electron is moving in the direction along with the current (as shown in figure) at a
speed of 0.01c, would it experience a magnetic force? If so, what is the direction and
magnitude of it? If not, why not? [c is speed of light = 3 x 108 m/s.]
c. Would the electron exert a magnetic force onto the wire? If so, what’s the direction and
magnitude of the force? If not, explain why not.
J
(a) I = J A = (8000)(π r2) = 13.2 mA
(
-
e
)
r µ I
4π ⋅10 −7 (0.0132 )
B= o =
= 5.28nT ⊕
2πr
2π (0.5)
0.5 m
v
r
r
(b) F = qvr × B = (− e )(0.01c )(5.28 ⋅10 −9 )sin 90o = 2.5 ⋅10 −21 N ↓
Electron moves away from the wire.
(c) Yes, action and reaction are equal and opposite.
So the electron is pushing the wire away (upward)
with the same force. 2.5 x 10-21 N.