A second proton with v=2.310^6 j hat m/s experiences F= 3.6910^16 k hat N in the same field.
What is the magnitude of B?
What is the direction B as an angle measured couter clock wise from the +x axis?
3 Answers

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

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

(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:

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

(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 / sin1(x) / cos1(x) ) to solve for θ


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