Question

2) The force that a magnetic field exerts on a charged particle is given by Ę = qö xĒ. A particle with mass m= 2.0x108 kg and charge q = +2.5x10-8C has an initial speed of v = 4+2 x 103 m/s (in the x- y plane). The magnetic field vector and velocity vector are B and û, respectively are displayed on the coordinate axis below. The angle between the vectors is 135 degrees. Use unit vector notation when describing the vectors. N (0.0, 10) (0-100) (0, 10, 0) Y 1100,0) х a) Write the Cartesian expression for the magnetic field vector: B = b) Write the Cartesian expression for the velocity vector: V = c) Write the Cartesian expression for the force on the charge: F = d) Find the magnitude of the force acting on the charge using the Cartesian vector approach: e) Calculate the magnitude of the force using: F = qvB sin ZOB)

Answer 1

a) Cartesian expression for the magnetic field vector: P = (109) 600ê in Tesla 1 x (b) = 452 x 103 m/s Angle between wand 3 is 135o and they are in X Y plane So e makes an angle of 45° and 45° withi-x) and y aseis respectively So, cartesian expression will be, 20|sin 45° ê + luy cos 45º 31 31 - 4x103 x + 4x 10 in mis y ego (e) F = qu x B² 2 x B = |- 4x108 4x10 <NO 3. 600 - 2.4 x 10 z 11 So, F = (2.5 x 1080) (-2.4 x 50 82) (-0.06 N) -

(d) Magnitude of the force, F = 160.06)2 + 02+02 0.06 N (e) Force, F = que sink(+9,8')) 29 = 10) √ - 4x103)² + (4x10²)2 5,66x103 mis 600T B = 161 ✓ (600)² +0² L(TXT) = 135° x Sin135 So, F = (5. 66x103) x 600 x 25 x 10 -0.06 N