A proton moves in the magnetic field B = 0.42i T with a speed of 1.0times10^7 m/

Question

A proton moves in the magnetic field B = 0.42i T with a speed of 1.0times10^7 m/s in the directions shown in the figure. (Figure 1) Part A In Figure (a), what is the magnetic force F on the proton? Give your answers in component form. Express vector F in the form of F_x, F_y, F_z, where the x, y, and z components are separated by commas. Part B In Figure (b), what is the magnetic force F on the proton? Give your answers in component form. Express vector F in the form of F_x, F_y, F_z, where the x, y, and z components are separated by commas.

Explanation / Answer

force=q*v*b*sin(theta)
for (a)
q = 1.6 x 10^-19;

v = 1 x 10^7;

B = 0.42 T along x axis;

theta=45

F = (1.6 x 10^-19)*(10^7)*(0.42)*sin(45) = 0.475 x 10^-12 N

now apply screw rule you will find the force is in y direction bring it to z and x componet by the formula force*cos(theta)..since angle between coordinate axis is 90 degree so there is no x or z component of force thus force is 0x + 0.475 x 10^-12y + 0z = 0.475 x 10^-12y

for (b)
theta =180 degree and all other are same
so force comes out to be zero there is thus no force Fx or Fy or Fz

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