Question

Q1. A beam of protons (q 1.6 x 10-19 C) moves at 3.0 x 105 m/s through a uniform 201 alongthe postive z-axis, as shown in the figure. The velocity of each proton lies in the xz-plane and is directed at 30° to the +z-axis. Find the force on a proton. Q2. The figure is a perspective view of a flat surface with area 3.0 cm2 in a uniform magnetic field B. The magnetic flux through this surface is +0.90 mWb. Find the magnitude of the magnetic field and the direction of area vector A -30 Q3. A particle with a charge of 1.24 x 10-8 C is moving with instantaneous velocity (4.19 x 104 ms)i(-3.85 x 104 ms 1)j. What is the force exerted on this particle by a magnetic field (a) B (1.40 T)t and (b) B (1.40T)R.

solve these 3 question step by step plz

Answer 1

Magnetic force exerting on the charge particle depends on the cross product of velocity and magnetic field. Here particularly the direction of magnetic field is different in both parts but magnitude is same so in the same scenario the direction of magnetic force is different in both parts

Magnetic force exerting on the charge particle depends on the cross product of velocity and magnetic field. Here particularly the direction of magnetic field is different in both parts but magnitude is same so in the same scenario the direction of magnetic force is different in both parts Given charge on particle q = -1.24 times 10^- 8 c v = {(4.19 times 10^4) i + (- 3.85 times 10^4) j } mol^- 1 (a) B = {(1.40)t} Tesla Magnetic force on moving particle F = q (V times B) = q Vertical-bar i 4.19 times 10^4 1.40 j - 3.85 times 10^4 0 k 0 0 Vertical-bar = q {i(0) - j(0) + k (- 3.85 times 10^4 times 1.40)} = -1.24 times 10^- 8 times (+ 3.85) times 10^4 times 1.40 k F = -6.683 times 10^- 4 kN in -ve z direction (b) if B = 1.40 T(k) implies F = q(v times B) = (q)