Presentation on theme: "Advanced Higher Physics Unit 2"— Presentation transcript:
1Advanced Higher Physics Unit 2 Motion in a magnetic field
2You need to be able to derive this! ForceA moving charge in a magnetic field experiences a force equal to:Since a moving charge isa current.In data bookletNow if:andThereforeYou need to be able to derive this!
3Notes: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 ableto derive this!In data bookletWith: 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).
4Direction of the force F B v The direction of the force is perpendicular to v and B.BFor a negative charge, use the right-handrule.FFvBelectronvThe force is out of the screen.
5Direction of the force v The direction of the force is perpendicular to v and B.BFor a positive charge, use the left-handrule.FvvPositive chargeThe force is into the screen.
6Direction of the force B When the velocity is NOT perpendicular to the magnetic field, then the componentof the velocity perpendicular to themagnetic field must be used:vsinθWhere θ is the angle between v and B.vsinθvθ
7ExampleCalculate the magnitude of the force on an electron travelling at, in a direction perpendicular to a magnetic field of 75 mT.B=75mTvElectron moving with a velocity of
9Path of a charged particle entering the field perpendicularly 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 turnchanges the direction of theforce and so on.The particle moves in a circle, andso the force is centripetal.Hence andEquating an cancelling gives:-qFrBvFv
10Example 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 fieldAfter leaving the magnetic field.
12Path of a charged particle entering the field at an angle. BThe velocity of the particle has acomponent parallel to the field (vcosθ)and a component perpendicular to thefield (vsinθ).vθvcosθvsinθ+qThe perpendicular component (vsinθ) causes circular motion asdescribed before.The parallel component is unaffected by the field and stay constant.Combining these two components produce helical motion.
13Applications of Electromagnetism Read green notes pageRead slide velocity selector and mass spectrometer fromVirtual AH Physics.
14JJ 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 fromVirtual AH Physics.Follow instructions from Virtual Experiment 5: e/m for anelectron.
15ExampleA 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 beyondthe 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.
16e) Calculate the electric field strength required to produce the proton path in part d).