A proton moving in a uniform magnetic field with v=1.18*10^6 m/s i hat experiences F=1.53*10^-16 N k hat?

3 Answers

• Let Vector-B, having magnitude "B" T make an angle θ counterclockwise with i hat.

  We have

  (Bsin θ)(1.1810^6)q = 1.53*10^-6, ------------------ 1 and

  (Bcos θ)(2.310^6)q = -3.69*10^-6;--------------------- 2squaring and adding

  (B^2)(1.6^2)(10^-38)(10^12){1.18^2 + 2.3^2} = (10^-12){1.53^2 +3.69^2} or

  B^2 = 0.9310^14 or B = 0.9710^7 or 9.7*10^ T

  Disregarding the sign dividing 1 by 2

  θ = tan^-1[0.8081] = 38.94 or 38.9 degree

• Let'slearntothink... more like let'sforgethowtoadd

  Using the formula F = (q)(v)(B)(sin θ) we find that:

  Let Vector-B, having magnitude "B" T make an angle θ counterclockwise with i hat

  1. (1.610^-19)(1.1810^6 m/s)(B)(sin θ) = 1.53*10^-16

  2. (1.610^-19)(2.310^6 m/s)(B)(cos θ) = -3.69*10^-16;

  solve for (B)(sin θ) and (B)(cos θ), we get:

  3. (B)(sin θ) = 1.5310^-16 / (1.610^-19)(1.1810^6 m/s) = 8.10410^-4

  4. (B)(cos θ) = -3.6910^-16 / (1.610^-19)(2.310^6 m/s) = 1.002710^-3

  squaring and adding the two results then applying trig identity (sin θ)^2 + (cos θ)^2 = 1

  (B^2)(sin^2 θ) + (B^2)(cos^2 θ) = (B^2)( (sin θ)^2 + (cos θ)^2) = B^2

  B^2 = (8.10410^-4)^2 + (1.002710^-3)^2 = 1.662*10^-6

  B = 1.289 mT

  plug in the value of B in equation 1 or 2 (use arcsin / arccos / sin-1(x) / cos-1(x) ) to solve for θ

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