Question

A charged particle (with q = 1.0 x 10-2 C) that is moving through a uniform magnetic field has a velocity (6.0 x 105 m/s) i + (4.0 x 105 m/s) k when it experiences a force F due to the magnetic field. If B = (-3.0 x 10-3T) i +(4.0 x 10-3T)j +(5.0 x 10-3 T) k , calculate the force, in N (in terms of unit vectors). Equation: F=qvxB F=-(3.0 N) i -(6.0 N)j + (7.0 N) k F=-(8.0 N) i – (42 N)j + (24N)k F=-(8.0 N)i - (21 N)j + (12 N) k F = (4.0 N) i (8.0 N)j -(4.0 N)k F = (-16 N) i – (42 N)j + (24N)k

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Answer 1

TLL The magnetic force Charged Particle of Charge a Pluced in a magnetic fields, moving with velocity ✓ is a F qWxB) . ginen, V = (6.0x10mis) i +(4.0-10% mys) & Ô- 3x1037) +(4x1021) f +(5-10-31) Ñ

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