Chapter 27: Magnetism

Question 1

unit of magnetic field

Answer 1

Tesla, massive amount

Question 2

why do we have to use f= q (v x B) for magnetic force and not just f = q E like for electirc

Answer 2

no magnetic monopoles, thus saying “q” ina mganetoc sense is not possible

Question 3

define B (vector) direction

Answer 3

the direction in which the north pole of a compass needle tends to point

Question 4

restate the concepts encoded in the equation for the magnetic force

Answer 4

F = q(vxB)
F = q (vBsinΘ)
- very much sign dependent, each wuatity sign change will effectively flip the directrion of the force
- force ⟂ B
- force ⟂ velocity of charge
- use RH rule to determine direction of force ( if B not ⟂ to v then you already no that there is not a maximum force),

Question 5

when does Fb = 0

Answer 5

• q is neutral
• sinΘ = 0,
  i.e. Θ = 0 : v //B
  Θ= 180 : v // -B

Question 6

when is Fb at a max

Answer 6

F = q vbsinΘ
sinΘ = 1
when Θ = 90

Question 7

when do objects experience circular motion

Answer 7

when force is always perpendicular to motion( or its velcoity or force ⟂ velocity , by RH rule)

only if the filed (B in this case) is uniform in time and space. then we know that F is cosntatn but direction is always chanigng,

thus Fb never does any work on the moving charge , F ⟂ motion always , thus cos90 = 0, thsu W = Fdl cos theta= fdl(0) = 0

Question 8

acceleration in circular motion gievne by

Answer 8

v^2 / R , think of nt coordinateds, centripital acceltration

Question 9

NII on our Fb in ciruclar motion yields what result for R and the consequences (NB remember, not on formula sheet)

Answer 9

|F| = m |a|
|q| v B = m (v^2/R)
R = mv / |q|B , note here that mv is linear momentum

Question 10

NII on our Fb in ciruclar motion yields R = mv / qB consequences (NB remember, not on formula sheet)

Answer 10

• if particle has greater initial momentum, i.e. greater mass greater vi, then it will have a larger radius.
• if magnetic field is larger, charge is larger quanity then it will have a smaller radius

Question 11

what is the expresison for angular speed ω and units

Answer 11

ω = v/ R

unit : radians per second

Question 12

angualr speed i .t.o q ,B ,m adn the implications

Answer 12

amnimulate NII on the eq of motion for to get
ω = (|q| B) / m

implications

• larger B or q means msmaller radius form radius equation, but it does have a greater ω. this obeys conservation of momentum, has we have smae Pinitial,
• larger m means smaller radius faster speed
• smaller m, larger radius, slower speed

Question 13

frequency relation to ω and incl T

Answer 13

ω = 2π f, with f = 1/T

T is period pf circular path, time to go round 2π

Question 14

what motion will we get if V is not perpedicular to B

Answer 14

helical motion, v gets split up into V parrallel and v perpendicular, then v parallel component amounts to 0 and we get result, F = q (v⟂) B , so we are still movign in circular motion but are also moving with the same orginial vi in the forward direction, thsu we geta helical path

Question 15

Lorentz force

Answer 15

Fnet = F elec + F mag
Fnet = qE + q(vxB)

Question 16

does magnetic force act on stationary objects

Answer 16

no
only charges
only is they are moving
only is B is present
hence, F = q(vxB)

Question 17

results from Oersted Danish scientist

Answer 17

First evidence of relationship between magnetism and moving charge (eletrical
current)

when there is no current, the comapss needle points north ( probs due to earths magnetic currentP)

when there is a current present, then the needle delfects acc the current direction

Question 18

Magnetic interaction is described in two steps:

Answer 18

1. a moving charge generates a magnetic firdl B

2. that magneitc field B exerts a force on any ptoher moving charge present int he field

Question 19

in a straight current carrying wire, what is the magnitude of the net force due to a magnetic field ⟂ to the direction of the current flow

Answer 19

Fnet = I l B (current x length x magfield)

Question 20

in a straight current carrying wire, what is the magnitude of the net force due to a magnetic field not always ⟂ to the direction of the current flow

Answer 20

vector :
Fnet = I (LxB)
Fnet = I LBsin Θ

Question 21

Definition of J , current density, vector J

Answer 21

current per unit charge
J = I/A
J= nqV

Question 22

change in kinetic energy due to B

Answer 22

0
W=0, by work energy theorem
B ⟂ v always

Question 23

in a curved current carrying wire, what is the magnitude of the net force due to a magnetic field not always ⟂ to the direction of the current flow

Answer 23

vector eq:

Fnet = ∫ dF = ∫ I (dLXB)

Question 24

when is the net force due to B zero

Answer 24

in a cicruclar current loop where the B is uniform

B ⟂ v always

Question 25

Torque eq

Answer 25

vectors: T = r x F = rFsinΘ= (distance from rotaitonal axis to position where force is being applied) x force

Question 26

what is μ

Answer 26

mu is the magnetic dipole defiend as
μ = I A ( where μ and A are vectors)
RH rule, curl fingers in direction of current and thumb points ot directions of unit vector μ hat

Question 27

Torque in current carrying loop i.t.o μ

Answer 27

T= μ x B (all vectors)

Question 28

potential energy i.t.o ϕ in current carrying loop

Answer 28

U(ϕ) = -μ·B = -μBcosϕ with U as a scalar, μ and B as vectors, duh

Question 29

when is a current carrying loop at a stable equilibirum i.t.o ϕ

Answer 29

when U is at a miniumum (a potential well)
shape of U vs ϕ graph is gpoevrned by eq U(ϕ) = -μBcosϕ .
thus min where cosϕ = -1, i.e where ϕ = 0 (or 360 which is basically 0)

where ϕ = 0

Question 30

when is a current carrying loop at an unstable equilibirum i.t.o ϕ

Answer 30

• when U is at a maximum ( a potential ‘hill’ of sorts)
• when cosϕ is at 1, U(ϕ) = -μBcosϕ

thus ϕ = 180

Question 31

when is a current carrying loop at a stable equilibirum NOT i.t.o ϕ but just interms of B and μ directions

Answer 31

a current loop which is free to orientate itself will
minimize its potential energy so that magnetic moment is parallel to the magnetic field.

i.e. ϕ = 0 and cos ϕ = -1 and U( ϕ ) = μB Joules. the potential energy well of the -cosϕ graph