A proton traveling at 4.40 km/s suddenly enters a uniform magnetic field of 0.74

Question

A proton traveling at 4.40 km/s suddenly enters a uniform magnetic field of 0.740 T , traveling at an angle of 55.0 o with the field lines (see the figure (Figure 1)).

1. Find the magnitude of the force this magnetic field exerts on the proton. F = ____ N

2. If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve. Fmax=?, Fmin=?

3. What would the answers to #1 be if the proton were replaced by an electron traveling in the same way as the proton?

Figure: 1 of 1

Explanation / Answer

Given

speed v = 4.4 km/s = 4.4 * 1000 m/s

v = 4400 m/s

magnetic field B = 0.74 T

angle   theta = 55

charge of proton is q = 1.6 * 10^-19 C

1)

The magnetic force on proton is F = q * v * B * sin(theta)

F = 1.6 * 10^-19 * 4400 * 0.74 * sin(55)

F = 4.27 * 10^-16 N

2)

from F = q * v * B * sin(theta)

Here F becomes maximum when sin(theta) = 1

that is theta = 90

Fmax = q * v * B

Fmax = 1.6 * 10^-19 * 4400 * 0.74

Fmax = 5.21 * 10^-16 N

Here F becomes minimum when sin(theta) = - 1

that is theta = 270

Fmin = - q * v * B

Fmin = - 1.6 * 10^-19 * 4400 * 0.74

Fmin = - 5.21 * 10^-16 N

3)

when proton is replaced by electron then there is no change in magnitude of force

because proton and electron has same charge

F = 4.27 * 10^-16 N (opposite to the direction of force in problem 1)

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