Question

At one instant, v = (-2.39^i + 2.38^j - 5.12^k) m/s is the velocity of a proton in a uniform magnetic field B = (2.93^i - 3.96^j + 9.47^k) mT. At that instant what are the x, y, and z components of the magnetic force F on the proton? What are the angle between V and F and the angle between v and B?

Answer 1

elocity of the proton,-239i+238j-5.12k)ms Magnetic field. B = ( 2.931-3.96/19.47k)×10% T Charge of the proton, q 1.6 x 10-C Magnetic force is given by, 19 =1.6×10-19 ×1-239 2.38 2.93x103.96x10 9.47 x10 5.12 -1.6x10-19 x2.26x10-31+7.63x10-3j+2.49x10-3k = (3.6x 10-221+1.22x 10-21j+3.98x10-22 k) N a) x- component of magnetic force, F-3.6x10-2N b) v-component of magneti c force.

-21 F, 1.22x10N c) z - component of magnetic force, F, 398x1022N d)

Angle between velocity and magnetic force, -(-2.391+2.38j-5.12K).(3.6x101+1.22x101j+3.98x10k -8.6 x1022 +2.9x10-22.04x10-21 Hence the angle, = cos-1 (0) = 90° f) Angle between velocity and magnetic field, F=qvB sin θ θ = sin qvB 1.33x10-21 - Sin 1.6x10-19 x 6.13x0.0106 -7.360