Problem

A proton moving at right angles to a magnetic field of $0.1 \, \mathrm{T}$ experiences a force of $2.0 \times 10^{-12} \, \mathrm{N}$. Determine the speed of the proton.

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Data

• Magnetic field: $B = 0.1 \, \mathrm{T}$,
• Magnetic force: $F_B = 2.0 \times 10^{-12} \, \mathrm{N}$,
• Charge of proton: $q = 1.6 \times 10^{-19} \, \mathrm{C}$,
• Angle between velocity and magnetic field: $\theta = 90^\circ$.

To find:

• Speed of the proton: $v = ?$.

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Prerequisite Concepts

The magnetic force acting on a charged particle moving in a magnetic field is given by:

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

Rearranging for $v$:

$$v = \frac{F_B}{q B \sin \theta}$$

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Solution

1. Substitute the given values:

$$v = \frac{2.0 \times 10^{-12}}{1.6 \times 10^{-19} \cdot 0.1 \cdot \sin 90^\circ}$$

2. Simplify step-by-step:
   • Calculate the denominator:

$$1.6 \times 10^{-19} \cdot 0.1 \cdot \sin 90^\circ = 1.6 \times 10^{-20}$$

• Divide the values:

$$v = \frac{2.0 \times 10^{-12}}{1.6 \times 10^{-20}} = 1.25 \times 10^{8} \, \mathrm{m/s}$$

3. Round off:

$$v \approx 1.3 \times 10^{8} \, \mathrm{m/s}$$

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Answer

The speed of the proton is approximately $1.3 \times 10^{8} \, \mathrm{m/s}$.