Question

A particle with charge of 2 (HC), mass of 2x10-20 (kg) and velocity of 200 (m/s) enters into a magnetic field of 0.06 (T). The particle experiences a force of 1.01x105 (N) in the magnetic field. A) Calculate the angle between velocity and magnetic field directions. B) Calculate the acceleration of charged particle in the magnetic field. C) Draw a diagram showing direction of velocity, magnetic field, force and the angle between velocity and magnetic field, using RHR-1 D) In order to maximize the force, what should be the angle between velocity vector and magnetic field direction? What would be the value of the maximum magnetic force at this angle?

Answer 1

A) solution. Given that 9 = 2MC= 2x10-6C; B= 0.06 T 2x10-20 kg fa lolx10-5N y = 2oomIS As, the magnetic force. F- EVB sino where "o" is the angle between velocity and field directions magnetic 101410-5 L .. sino = 2413 2x10-G x 200 x 0.06 sino = 0.04208 xlo Or sinn (o. 4208) 24.885 13) Solutions since 9113sino. According to "newton's second low" f = ma quBsino ma quissino 2x1076x 200 x0.06x sin (24.885) 2x 10-20 las 5.04958 1014 m/sec2. CS Scanned with CamScanner

c) Solutions 2 TU laby P 900 Magnetic force velocity (T? B Gragnetic field) Here Thumb finger represent & Magnetic force (F) Index finger represents velocity () Middle finger represents - Magnetic field (B E = D since, Eub sino "F" becomes maximum, when sino=1 then 0=90 F- PUB F = 2x10-6x200 x0.06 : 2.4x10 -IN CS Scanned with CamScanner