A charge q = 1.008E-17 C moves with velocity v = v_z k = 8500 k m/s in a uniform

Question

A charge q = 1.008E-17 C moves with velocity v = v_z k = 8500 k m/s in a uniform electric field E = = E_x i = 95 i V/m and a uniform magnetic field B = B_y j = 0.75 j T. Refer to the figure. Randomized Variables q = 1.008E-17 C V = V_2K = 8500 K m/s E = E_xi = 95 i V/m B = B_y j = 0.75 j T (a) Express the magnitude of the electric force acting on the charge F_E. in terms of E_x and q. (b) Calculate the value of FE. in newtons. (c) What is the direction of the electric force on the charge? (d) Express the magnitude of the magnetic force. F_B. acting on the charge in terms of B_y, V_z and q. (e) Calculate the value of F_B. in newtons. (f) What is the direction of the magnetic force on the charge?

Explanation / Answer

Here,

part a)

FE = Ex * q

part b)

FE = 95 i * 1.008 *10^-17 N

FE = 9.576 *10^-16 i N

the value of FE is 9.576 *10^-16 N

part c)

as FE = 9.576 *10^-16 i N

the direction of force on the charge is + ve x direction

part d)

for the magnetic force

magnetic force = q * Vz X By

magnetic force = q * Vz * By

the magnitude of magnetic force is q * Vz * By

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