A proton traveling at 4.50km/s suddenly enters a uniform magnetic field of 0.760

Question

A proton traveling at 4.50km/s suddenly enters a uniform magnetic field of 0.760T , traveling at an angle of 55.0o with the field lines (see the figure(Figure 1) ).

A) Find the magnitude of the force this magnetic field exerts on the proton.

B) If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve.

C) What would the answers to part (B) be if the proton were replaced by an electron traveling in the same way as the proton?

A proton traveling at 4.50km/s suddenly enters a uniform magnetic field of 0.760T , traveling at an angle of 55.0o with the field lines (see the figure(Figure 1) ). A) Find the magnitude of the force this magnetic field exerts on the proton. B) If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve. C) What would the answers to part (B) be if the proton were replaced by an electron traveling in the same way as the proton?

Explanation / Answer

A.
F = q v B sin (theta)
= (1.6 x 10^-19 C) (4.5x10^3 m/s) (0.760 T) sin (55*pi/180 rad)
= 4.482 x 10^-16 N
Direction: Force vector is perpendicular to the plane formed by the velocity vector and magnetic field vector, in the direction of the cross product of V and B vectors. Thus in the present case, force vector is in the direction pointing normally inside the screen (using right hand thumb rule).

B.

Maximum magnitude of force will be when sin (theta) is 1, ie theta is 90 degrees. This means that velocity vector is perpendicular to the magnetic field.

Fmax = q v B sin (90)
= (1.6 x 10^-19 C) (4.5x10^3 m/s) (0.760 T)

= 5.472 x 10^-16 N

Minimum magnitude of force is when sin (theta) magnitude is minimum, which is 0 when theta is 0 degrees or 180 degrees. This means that velocity vector is parallel or anti parallel to the magnetic field.

Fmin = q v B sin (0) = 0, parallel

Fmin = q v B sin (180) = 0, anti-parallel

c)

since photon and electron have same charge...the force will remain same for part b for proton also.

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