# Advanced Higher Physics Unit 2

## Presentation on theme: "Advanced Higher Physics Unit 2"— Presentation transcript:

Motion in a magnetic field

You need to be able to derive this!
Force A moving charge in a magnetic field experiences a force equal to: Since a moving charge is a current. In data booklet Now if: 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: You need to be able to derive this! In data booklet 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 F B v
The direction of the force is perpendicular to v and B. B For a negative charge, use the right-hand rule. F F v B electron v The force is out of the screen.

Direction of the force v
The direction of the force is perpendicular to v and B. B For a positive charge, use the left-hand rule. F v v Positive charge The force is into the screen.

Direction of the force B When the velocity is NOT perpendicular
to the magnetic field, then the component of the velocity perpendicular to the magnetic field must be used: vsinθ Where θ is the angle between v and B. vsinθ v θ

Example Calculate 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

Path of a charged particle entering the field perpendicularly
The particle experiences a force perpendicular to both B and v (right-hand rule if q negative). This force changes the direction of the particle which in turn changes the direction of the force and so on. The particle moves in a circle, and so the force is centripetal. Hence and Equating an cancelling gives: -q F r B v F v

Example A proton travelling at , enters a uniform magnetic
field, acting over circular area, along a diameter as shown. The proton leaves the magnetic field. Describe the velocity: Inside the magnetic field After leaving the magnetic field.

Path of a charged particle entering the field at an angle.
B The velocity of the particle has a component parallel to the field (vcosθ) and a component perpendicular to the field (vsinθ). v θ vcosθ 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 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.

Example A 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. Calculate the force acting on the proton inside the magnetic field. Calculate the radius of curvature of the proton path in the magnetic field. 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 the proton is undeflected. Show on a sketch how this field is applied showing the polarity of any electrodes.

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

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