PHYS1231 HIGHER PHYSICS 1B
Solutions Tutorial 3
Magnetic Field - Basic info:
Unless otherwise specified, the symbols E, F, l, v, r and B represent vector
quantities of electric field, force, length, velocity, radial displacement and
magnetic field intensity respectively. The force experienced by a charge, q
moving at a velocity, v through a magnetic field, B is given by:
F = qv ! B
The force experienced by a wire length, l carrying a current, I in a magnetic field
is given by:
F = il×B
The Right Hand Rule (RHR) is frequently used to gain an intuitive understanding
of the direction of these forces but remember that “i” describes the direction of
positive charge. Wrap your fingers in the direction from v to B. Then the force on
a positive charge is in the direction of your thumb. (For a negative charge it is in
the opposite direction.)
Others may find an easy way to
remember this:
The Biot-Savart Law describes the magnetic field generated at a displacement, r
from a current, i moving along some path dl.
µ idl # rˆ
! dB = 0 !
4"
r2
Note that rˆ is the unit vector in the direction of r with a magnitude = 1.
Ampere’s Law states that the integral dot product of a magnetic field around any
closed path will be proportional to the current passing through that enclosed loop.
! B.dl = µ 0 i
Questions:
1. (refer to diagram) E = 1500V/m, B = 0.4T and q = 1.6 X10-19C. The force on
the charge (1.6 X10-19C) due to the electric field FE = Eq = 2.4 X10-16N
downwards. For that force to be balanced with a magnetic one (and
assuming the velocity of the electron is ⊥ to the direction of both E and B), FB
= qvB = 6.4 X10-20 v = 2.4 X10-6N. Then v = 3750m/s
2. (a) From the RHR, the vertical down magnetic field and the horizontal S to N
velocity, the force acting on the NEGATIVELY charged electron will be West
to East. (b) First we need to get the velocity of the electron which we can find
from its kinetic energy = mev2/2= 12 KeV = 1.92 X 10-15J. Then v= 6.5X106m/s
(the mass of an electron is 9.11X10-31kg)… then the acceleration of the
electron is found by a = qvB/me = 6.3 X1014 m/s2 (c) Now, we just use linear
dynamics. It takes the electron 3.08 X10-8 s to travel the 0.2m South to North,
and in that time it is deflected West to East by a displacement s = ut +at2/2.
Here u = 0, so s =0.3m.
3. Force of gravity downwards = 0.01kg X 9.8m/s2 = 0.098N. To counterbalance
this we need 0.098N upwards. Therefore, with the field, B = 0.40T coming out
of the page, we need a current flowing from right to left with a magnitude =
0.098N/Bl = 0.41A
4. Path I has an enclosed current passing through it = i. The corresponding
magnetic field comes out of the page within the loop of the wire and into the
page on the other side. Path II has an net enclosed current of zero; the path
integral of the magnetic field around Path II will also = 0. Path III has an
enclosed current passing through it = i. The corresponding magnetic field
comes out of the page within the loop of the wire and into the page on the
other side.
5. At (a) the magnetic field is out of the page, at (b) it is into the page, at (c) it is
also into the page but the magnitude will be less than that at (b).
6. From Ampere’s Law we can derive an expression for B at a displacement, r
away from a long current carrying wire by integrating along a circular path
(2πr) around the wire.
2"
# B.rd! = µ0i
0
From which we get
B = µ0i/2πr
(a) If the electron is 0.05m away, and travelling ⊥ towards the wire, this magnetic
field produces a force parallel to the current in the wire = qv(µ0i/2πr) = 3.2X10-16N
(b) If the electron is at the same distance away and travelling parallel to the
direction of current, the same force as above will be directed radially out away
from the wire. (c) If the electron is ⊥ to both the above directions then it is
travelling either at 00 or 1800 with respect to the magnetic field and, from the
cross product F = qv ! B , the resultant force should = 0
7. Consider just the field due to the 30A in the long wire. At a distance “a” from
that wire Ba = µ0i/2πa = 6 X10-4T into the page. At a distance “b”,
Bb = µ0i/2π(a+b) = 6.67 X10-5T into the page. Note that the effect of the field
on the sides of the loop of length “b” is to produce equal but opposing forces.
For the side of the closed loop at “a”, the force on this length of wire carrying
20A is given by:
Fa = ilBa = 3.6 X10-3N to the left
Likewise
Fb = ilBb = 4.0 X10-4N to the right.
The net force Fa + Fb = 3.2 X10-3N to the left
INDUCTION AND MAGNETISM
Basic concepts
Flux: If we think of magnetic field lines as representing the location and intensity
of a magnetic field, then magnetic induction B = number of field lines passing
perpendicularly through a given area, A
! = B"A
Yes, that is a dot product. The "direction" of an area is taken here to be
perpendicular ⊥ to the surface, A pointing away in the general direction of B.
Faradays Law: A change in the magnetic flux through an area produces an emf.
emf =
d!
dA
dB
=B
+A
dt
dt
dt
(1)
Lenz's Law: The emf produced by a changing magnetic flux will act to oppose
that change in magnetic flux (magnetically speaking, Nature likes to keep the
status quo).
d#
(2)
emf = !1 "
dt
Self inductance: The quality of an inductor = "L". Often a real inductor is a
solenoid: a helical coil of wire. When a changing current (i) is run through this
inductor:
di
(3)
emf = !L
dt
…or if we think of that coil as having N loops or turns, each having a flux passing
through them, then combining (2) and (3)…
N! = Li
(4)
Mutual inductance: when you have two coils (1&2) arranged such that the flux
generated by one coil also passes through the second, then:
(5)
N 2 !2 = Mi1
Where M is the coefficient of mutual inductance (this situation occurs in electrical
transformers).
Energy in a magnetic field: in an analogous way to the energy stored in a
capacitor, the energy stored in an inductor (with inductance, L) is:
1
U = Li 2
(6)
2
Magnetic fields in matter
Where H = magnetic field strength (in A/m), µ0 = 4π X 10-7 Tm/A and M =
magnetisation in a material (again in A/m), the magnetic induction, B is given by:
B = µH = µ 0 (H + M )
and further… M= χH so B = µ 0 (1 + !)H or µ = µ0(1+ χ)
(χ = magnetic susceptibilty and µ = magnetic permeability.)
(7)
(8)
The magnetic dipole moment of a magnetised object, mx (along a given axis "x")
is related to that object's volume and magnetisation along the same axis, Mx by:
(9)
m x = M x ! Volume
Questions
8. (a) From Lenz's Law, the current in the smaller loop will be driven by an
induced emf that will oppose the change in magnetic flux. This change comes
from the sudden introduction of current into the larger loop causing an
increase in magnetic flux from L to R. Therefore, the field generated in the
smaller loop will be R to L and, from the right hand rule (RHR) the current will
be flowing anticlockwise. (b) We might intuitively say that the B induced the
small loop will be opposite that in the large and so there will be repulsion.
However the ultimate way to calculate that force is using F = I l × B and the
real situation involves some tricky thinking about the geometry of the
magnetic field from the large coil as it meets the small coil - see diagram:
9. (a) Right after the switch is closed, there will be a current running through the
solenoid such that an increasing magnetic field will be produced flowing from
right (r) to left (L). By Lenz's law there will arise an emf and hence a current in
the coil connected to the resistor, R such that an opposing magnetic field is
induced flowing L - r. This requires a current through R flowing from r - L. (b)
After the switch has been closed for a while there will be a stable, unchanging
magnetic field and, therefore, no induced emf and no current. (c) Immediately
after the switch is open again the flow of current through the solenoid will
decrease and there will be a corresponding decreasing magnetic field. This
will induce an emf in the coil connected to R that will seek to oppose this
change by pushing a current through R from L - r. (d) the left end, as the field
lines emerge from this end.
10. (a) When the rod moves to the left with v = 8 m/s the change in magnetic flux
can be found through:
d!
dA
dB
emf =
=B
+A
(10)
dt
dt
dt
Note that the magnetic field, B = 1.0T doesn't change with time, so the last
term in the above expression = 0. dA/dt = 8 m/s X 0.5m = 4m2/s… and so:
dA
emf = B
= 4V
(11)
dt
The change in the amount of flux through the ABCD loop is downwards, so
the direction of the emf will oppose this - pushing a current counter-clockwise
around the loop. (b) If there is 4V induced around a circuit of resistance = 0.4
Ohms, then the current i = 10A. Then, by
F = il ! B
(12)
The force on this current carrying section = 10A X 0.5m X 1.0T X sin (900) to
the right. The force required to maintain constant motion = 5N to the right. (c)
The rate of mechanical work done, dW/dt = F. ds/dt = F.v = 40 Watts. The
rate of electrical power dissipated, P = Vi = 40 Watts.
11. (a) The flux moving through this loop Φ = B.A = BNabcos(θ). Although B is
constant, θ is constantly changing with time. As the loop turns around θ = ωt =
2πνt. Hence,
d#
emf = $
= 2!"BNab sin( 2!"t )
(13)
dt
(b) Where ν = 60Hz, B = 0.5T and where we want the maximum
emf ε0 = 150V, then 2πνBNA = 150V (A = ab). Therefore
NA = 5/2π turns/metre2.
12. For a long solenoid carrying a current i, the magnetic field B along the centre
= µ0ni, where n = the number of turns per metre. For the case here the total
number of turns N = 2m ÷ 0.00252 = 793.65 because the adjacent turns are
touching. Then in 1m n = 396.83 m-1 and, with Φ = B.A and A = π (0.0205m)2,
Φ = 0.524µ0i. From
(14)
N! = Li
-4
We can say L = N X 0.524µ0 = 5.226 X 10 Henries. Now, to determine the
inductance/metre divide by 2m: L/m = 261 µH/m.
13. The number of turns/metre in this solenoid, n = 1870/1.26m = 1484 m-1, and
using similar reasoning to Q 5. See that L = NAµ0µn = An2µ0µl. Then L = (π X
0.027252) X (1870)2 X (4πX10-7) X 968 X 1.26m = 7.87H
14. (a) Let us say that the emf across one inductor is emf1 and across a second,
in series, is emf2. The relationship between these emfs and their respective
inductances is given by:
d
d
emf 1 = L1 I 1 and emf 2 = L2 I2
dt
dt
Let us also say that the combined emf across these inductors is emfT. Now, if L1
and L2 are in series then the current through both is the same I1 = I2 = I. Hence if
we describe emfT by a combined inductance LT, also with current = I,
dI
dI
dI
emf T = LT = L1 + L2
or LT = L1 + L2
dt
dt
dt
(b) Referring to the diagram below…the junction rule tells us that if we have a
parallel connection then the sum of the currents running through each branch (I2
+ I3) = I1. However, the voltage or emf across both the inductors shown is equal.
Then, the same emf across a single inductor L, that would represent the parallel
combination of L1 and L2, would be defined by emf = -Leq.(dI/dt). Or if you like…
! emf
L eq
So if
=
dI
dt
but note that
! emf
L1
=
d I2
! emf d I 3
and
also.
=
dt
dt
L2
! emf ! emf ! emf
dI d (I 2 + I 3)
then
or
=
=
+
dt
dt
L eq
L1
L2
1
L eq
=
1
L1
+
1
L2
(15)
.
The inductors are assumed to be separated by a large distance so that their
magnetic fields do not affect each other. ie. there's no mutual inductance.
15. (a) Using the following equation with i = 62.0 X 10-3A and U = 25.3mJ…
1
U = Li 2
(16)
2
Therefore L = 13.2 H. (b) To quadruple the magnetic energy you need to double
the current to i = 124mA