Advanced Higher Physics Unit 2

Motion in a magnetic field

Force

sinBIlF

t

qI vtdl

A moving charge in a magnetic field experiences a force equal to:

Since a moving charge isa current.

In data booklet

Now if:

sinvtt

qBF

sinqvBF

and

Therefore

You need to be able to derive this!

Notes:

•This shows that a charge must be moving to experience a force and that the movement must not be parallel to the field.

•The most common case is when a charge is moving perpendicular to the field (θ=90˚, sin90=1). In this case the equation reduces to:

qvBF In data bookletYou need to be able

to derive this!

With: F magnitude of the force in N.q magnitude of the charge in C.v velocity of the charge in msˉ¹.B magnetic induction in T (B is perpendicular to v).

Direction of the force

B

v

electron

The direction of the force is perpendicular to v and B.

For a negative charge, use the right-handrule.

v

B

F

The force is out of the screen.

F

Direction of the force

B

v

Positive charge

The direction of the force is perpendicular to v and B.

For a positive charge, use the left-handrule.

The force is into the screen.

F

v

Direction of the force

B

v

When the velocity is NOT perpendicularto the magnetic field, then the componentof the velocity perpendicular to themagnetic field must be used:

vsinθ

Where θ is the angle between v and B.

θ

vsinθ

Example

16105.4 msCalculate the magnitude of the force on an electron travelling at , in a direction perpendicular to a magnetic field of 75 mT.

B=75mT

v

Electron moving with a velocity of 16105.4 ms

Path of a charged particle entering the field perpendicularly

qvBF r

mvF

2

B-q

v

The particle experiences a forceperpendicular to both B and v(right-hand rule if q negative).This force changes the directionof the particle which in turn changes the direction of the force and so on.The particle moves in a circle, andso the force is centripetal.Hence and

Equating an cancelling gives:

F

v

F

r

mvqB

r

Example

A proton travelling at , enters a uniform magneticfield, acting over circular area, along a diameter as shown.

The proton leaves the magnetic field.Describe the velocity:(a)Inside the magnetic field(b)After leaving the magnetic field.

16105.2 ms

Path of a charged particle entering the field at an angle.

B

v θ

vsinθ

vcosθ

The velocity of the particle has a component parallel to the field (vcosθ)and a component perpendicular to the field (vsinθ).

+q

The perpendicular component (vsinθ) causes circular motion as described before.

The parallel component is unaffected by the field and stay constant.

Combining these two components produce helical motion.

Applications of Electromagnetism

•Read green notes page 24-26.•Read slide velocity selector and mass spectrometer from Virtual AH Physics.

JJ Thomson’s experiment to measure the Charge to Mass ratio for electrons

•Read green notes page 26.•Read slide charge to mass ratio for an electron from Virtual AH Physics.•Follow instructions from Virtual Experiment 5: e/m for an electron.

ExampleA proton travelling at enters a uniform magnetic field 65 mT. The magnetic field is perpendicular to the velocity direction, into the screen, as shown in the diagram.

a) Calculate the force acting on the proton inside the magnetic field.b) Calculate the radius of curvature of the proton path in the magnetic field.c) Describe and draw a sketch to show the path of the proton in and beyond the magnetic field.d) A uniform electric field is applied and adjusted so that the path of theproton is undeflected. Show on a sketch how this field is applied showing thepolarity of any electrodes.

16102.3 ms

e) Calculate the electric field strength required to produce the proton path in part d).