Question

A α-particle moves in a circle of radius 5.0*10^-2m in magnetic field of 2.0T. Find the speed of the α-particle.

Answer 1

Answer on Question 75491, Physics, Electromagnetism

Question:

A $\alpha$-particle moves in a circle of radius $5.0 \cdot 10^{-2} \, \text{m}$ in magnetic field of $2.0 \, \text{T}$. Find the speed of the $\alpha$-particle.

Solution:

There are two forces that act on the $\alpha$-particle when it moves in the uniform magnetic field: the magnetic force and the radial force. So, using the Newton's second law of motion we can write:

$$q v B = \frac{m v^2}{r},$$

here, $q$ is the charge of the $\alpha$-particle, $v$ is the orbital speed of the $\alpha$-particle, $B$ is the magnetic field, $m$ is the mass of the $\alpha$-particle, $r$ is the radius of the curvature of $\alpha$-particle's path.

From this formula, we can find the orbital speed of the $\alpha$-particle:

$$v = \frac{q B r}{m} = \frac{2 \cdot 1.6 \cdot 10^{-19} \, C \cdot 2.0 \, T \cdot 5.0 \cdot 10^{-2} \, m}{6.644 \cdot 10^{-27} \, kg} = 4.82 \cdot 10^6 \, \frac{\text{m}}{\text{s}}.$$

Answer:

$$v = 4.82 \cdot 10^6 \, \frac{\text{m}}{\text{s}}.$$

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