A particle with positive charge q = 3.20 10-19 C moves with a velocity v with ar

Question

A particle with positive charge q = 3.20 10-19 C moves with a velocity v with arrow = (2i + 3j k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving particle (in unit-vector notation), taking B with arrow = (2i + 4j + k) T and E with arrow = (4i j 2k) V/m. (a) After substituting and combining the resulting coefficients according to the rules above, we have the following. F = 3.2 × 1019 C i j + k N/C = i j + k × 1018 N

Explanation / Answer

magnetic force Fb = q*(v x B)

Fb = 3.2*10^-19*( (2i + 3j - k) x (2i + 4j + k) )

Fb = 3.2*10^-19* ( 7i - 4j + 2k )

electric force Fe = q*E = 3.2*10^-19*(4i - j - 2k )

total force F = Fb + Fe

F = 3.2*10^-19*( 11i - 5j )

F = 35.2*10^-19 i N - 16*10^-19 j N

angle made by the force vector with sx axis

theta = tan^-1(Fy/Fx)

theta = tan^-1( 16*10^-19/35.2*10^-19) = 24.4 degrees

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