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6) An electron has a velocity of 6.0 x10 m/s in the positve x direction at...
A negative charge q = −3.80×10^−6 C is located at the origin and has velocity υ⃗ =(7.50×10^4m/s)ι^+((−4.90)×10^4m/s)j^. At this instant what is the magnetic field produced by this charge at the point x = 0.210 m , y = -0.350 m , z = 0? Enter the x, y, and z components of the magnetic field separated by commas. Bx, By, Bz = μT
An electron that has a velocity with x component 2.0 x 106 m/s and y component 2.9 x 106 m/s moves through a uniform magnetic field with x component 0.032 T and y component -0.19 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity. answer must be in N units!!!!!!
An electron that has a velocity with x component 2.7 × 106 m/s and y component 2.3 × 106 m/s moves through a uniform magnetic field with x component 0.033 T and y component -0.21 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.
An electron (me 9.11 x 10-31 kg) has a speed of 4 x 106 m/s. If the uncertainty in its energy is 8 ev, what is the uncertainty in its position? O 1.65 x 10-10 m 2.11 x 10-10 m 2.64 x 10-10 m 3.30 x 10-10 m 2.07 X10-9
1) A beam of electrons with velocity of 9.0 x 10 m/s in the horizontal direction pass through a vertical E-field with a magnitude of E= 1500 N/C and impacts a detection screen, producing a glowing spot on the screen after crossing a horizontal distance of Ax 15 cm through the wiform E-field. (Electrons have a mass of 9.11 x 10-31 k.) electron beam V detection screen a) (5 points) Neglecting the effects of gravity, find the vertical acceleration a,...
An electron (charge e) with a velocity described by the vector v=(2.0, 3.5, 5.5) m/s at an instant in time is traveling through a uniform magnetic field of 6.0 T in the z-direction. Calculate the components of the magnetic force vector on the electron. A. Fx= 21e Fy = -12e Fz= 0 B. Fx= 21e Fy = -12e Fz= 33e C. Fx= 21e Fy = 12e Fz= 0 D. Fx= 12e Fy = 21e Fz= 0
An electron that has velocity (3.4 x 10 m/s)i+ (2.5 x 106 m/s)j moves through a magnetic field B (0.03 Ti (0.15 T)]. (a) Find the force on the electron magnitude X* N direction Etera mumber (b) Repeat your calculation for a proton having the same velocity magnitude direction
A cosmic ray electron moves at 7.4 x 106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.32 10-5 T. What is the radius (in meters) of the circular path the electron follows? The charge of an electron is 1.6 x 10-19 C and its mass is 9.11 x 10-31m. Your answer should be a number with two decimal places, do not include the unit.
A cosmic ray electron moves at 6.6 x 106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.37 10-5 T. What is the radius (in meters) of the circular path the electron follows? The charge of an electron is 1.6 x 10-19 C and its mass is 9.11 x 10-31m. Your answer should be a number with two decimal places, do not include the unit.
An electron that has velocity v=(3x10^5 m/s) and moves along positive x direction through the uniform magnetic field B=(0.8T) which is along the positive z direction. (a) Find the force on the electron (magnitude and direction). (e=1.6x10^-31 kg) (b) Calculate the radius of the electron's path in the magnetic field.