Problem

An $8 \, \mathrm{MeV}$ proton enters perpendicularly into a uniform magnetic field of $2.5 \, \mathrm{T}$.

(a) What is the magnetic force acting on the proton?
(b) What will be the radius of the path of the proton?

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Data

• Magnetic field: $B = 2.5 \, \mathrm{T}$,
• Charge of the proton: $q = 1.6 \times 10^{-19} \, \mathrm{C}$,
• Mass of the proton: $m = 1.67 \times 10^{-27} \, \mathrm{kg}$,
• Kinetic energy of the proton:

$$KE = 8 \, \mathrm{MeV} = 8 \times 10^6 \, \mathrm{eV} = 8 \times 10^6 \times 1.6 \times 10^{-19} \, \mathrm{J} = 1.28 \times 10^{-12} \, \mathrm{J}.$$

To find:
(a) Magnetic force: $F_B = ?$,
(b) Radius of the path: $r = ?$.

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Prerequisite Concepts

1. Magnetic Force: The magnetic force acting on a moving charged particle is:

$$F_B = q v B \sin \theta$$

Since the particle enters perpendicularly, $\sin \theta = 1$, so:

$$F_B = q v B$$

2. Kinetic Energy and Velocity: The velocity of the proton can be calculated using the relation between kinetic energy and velocity:

$$KE = \frac{1}{2} m v^2$$

Rearranging for $v$:

$$v = \sqrt{\frac{2 KE}{m}}$$

3. Radius of Circular Path: The radius of the circular path for a charged particle in a magnetic field is given by:

$$r = \frac{m v}{q B}$$

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Solution

(a) Magnetic Force:

1. Calculate Velocity:
   Using the formula for velocity:

$$v = \sqrt{\frac{2 KE}{m}}$$

Substituting the values:

$$v = \sqrt{\frac{2 \times 1.28 \times 10^{-12}}{1.67 \times 10^{-27}}}$$

Simplify:

$$v = \sqrt{\frac{2.56 \times 10^{-12}}{1.67 \times 10^{-27}}} = \sqrt{1.53 \times 10^{15}} = 1.24 \times 10^7 \, \mathrm{m/s}.$$

2. Calculate Magnetic Force:
   Using the formula:

$$F_B = q v B$$

Substituting the values:

$$F_B = (1.6 \times 10^{-19}) \cdot (1.24 \times 10^7) \cdot (2.5)$$

Simplify:

$$F_B = 4.96 \times 10^{-12} \, \mathrm{N}.$$

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(b) Radius of the Path:

1. Using the formula for the radius:

$$r = \frac{m v}{q B}$$

Substituting the values:

$$r = \frac{(1.67 \times 10^{-27}) \cdot (1.24 \times 10^7)}{(1.6 \times 10^{-19}) \cdot (2.5)}$$

Simplify the numerator and denominator:

$$r = \frac{2.07 \times 10^{-20}}{4.0 \times 10^{-19}} = 0.0518 \, \mathrm{m}.$$

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Answer

(a) The magnetic force acting on the proton is:

$$F_B = 4.96 \times 10^{-12} \, \mathrm{N}.$$

(b) The radius of the proton’s path is:

$$r = 0.0518 \, \mathrm{m} \, \text{or} \, 5.18 \, \mathrm{cm}.$$