Q1. The diagram below shows a velocity selector. protons travelling between the 2 charged plates experience a force due to the electric field (E). A magnetic field B causes a force on the protons in the opposite direction. The protons are moving in a straight line between the 2 plates. +VD Figure 1 (a) What direction is the force on the electrons due to the charged plates (1 mark) tris 300 V. and they are 10
0 sl Q Search Q4. (a) A An electron enters a velocity selector that has a potential difference (p. d.) across its plates of 2000 V and a magnetic flux density of 0.064 T. The plate separation is 0.04 m. Write down an expression for the force acting on the electron due to (i) The electric field (1 mark) (ii) The magnetic field (1 mark) (iii) Calculate the speed of the electrons which pass straight through the selector (1 mark) (b) A proton with kinetic energy 7 x 10-16 J moves perpendicular to a magnetic field of 0.34 T. What is the radius of its circular path? What is the time period for this circular motion? (The proton mass is 1.67 x 10-27 kg) (2 marks) (b) A wire which is carrying a current of 6.0 A has 0.5 m of its length placed in a magnetic field of size 0.2 T. Calculate the size of the force on the wire if it is placed: at right angles to the direction of the field; (1 mark) • (1 mark) at 45 degrees to the direction of the field; and along the direction of the field. ● (1 mark) > D
Q1. The diagram below shows a velocity selector. protons travelling between the 2 charged plates experience a force due
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Q1. The diagram below shows a velocity selector. protons travelling between the 2 charged plates experience a force due
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