Question

An alpha particle (a He nucleus containing two protons and two neutrons and having a mass of $\mathbf{6}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{kg}$) travelling horizontally at $\mathbf{35}\mathbf{.}\mathbf{6}\mathbf{km}\mathbf{/}\mathbf{s}$enter a uniform, vertical,$\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{-}\mathbf{T}$ magnetic field.(a) What is the diameter of the path followed by this alpha particle? (b) what effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) explain why the speed of the particle does not change even though an unbalanced external force acts on it.

Short answer

(a) The diameter of the path followed by this alpha particle is$8.208\times {10}^{-4}\mathrm{m}.$

(b) There is no any effect of magnetic field on the speed of the particle it means constant.

(c) The magnitude is $3.09\times {10}^{12}\mathrm{m}/{\mathrm{s}}^2$and the direction is radially and horizontally inward the curvature of the particle’s path.

(d) The velocity is perpendicular the force. It means magnitude is not changed and direction of the velocity should be change

Step-by-step solution

Definition of the magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the Diameter

The force acting on it

${\stackrel{\rightharpoonup }{F}}_B=q\left(\stackrel{\rightharpoonup }{v_0}\times \stackrel{\rightharpoonup }{B}\right)$

Velocity and magnetic field perpendicular to each other so

${\stackrel{\rightharpoonup }{F}}_B=\left|q\right|\left(\stackrel{\rightharpoonup }{v_0}\times \stackrel{\rightharpoonup }{B}\right)$ ...... (1)

And according to Second law of motion the magnitude of the force is product of mass and acceleration

${\mathrm{F}}_{\mathrm{B}}=\mathrm{ma}$

Here m is the mass of proton and a is the acceleration.

And radial acceleration is $\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$so

${\mathrm{F}}_{\mathrm{B}}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$ ..…(2)

Now form (1) equation $\stackrel{\rightharpoonup }{{\mathrm{F}}_{\mathrm{B}}}=\left|\mathrm{q}\right|\mathrm{vBsin}\left(\mathrm{\theta }\right)$where $\mathrm{\theta }=90°\mathrm{So}{\mathrm{F}}_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vB}$

From (1) and (2)

$$\begin{gathered} \left|\mathrm{q}\right|\mathrm{vB}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ \mathrm{R}=\frac{\mathrm{mv}}{\left|\mathrm{q}\right|\mathrm{B}} \end{gathered}$$

…

Charge on the alpha particle

$$\begin{gathered} \mathrm{q}=+2\mathrm{e} \\ \mathrm{q}=3.20\times {10}^{-19}\mathrm{C} \end{gathered}$$

Put all the values in

$$\begin{gathered} \mathrm{R}=\frac{\left(6.64\times {10}^{-27}\mathrm{kg}\right)\left(35.6\times {10}^3\mathrm{m}/\mathrm{s}\right)}{\left(3.20\times {10}^{-19}\mathrm{C}\right)\left(1.80\mathrm{T}\right)} \\ \mathrm{R}=4.104\times {10}^{-4}\mathrm{m} \end{gathered}$$

Diameter is twice of radius so

$$\begin{gathered} \mathrm{D}=2\mathrm{R} \\ \mathrm{D}=2\left(4.104\times {10}^{-4}\mathrm{m}\right) \\ \mathrm{D}=8.208\times {10}^{-4}\mathrm{m} \end{gathered}$$

Hence the diameter of the path followed by this alpha particle is$8.208\times {10}^{-4}m.$

Determine the effect of magnetic field on the speed of particles and determine the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field

On the particle acting force is perpendicular to the direction of motion so

$$\begin{gathered} W=\int \hat{F}.d\hat{s} \\ W=0 \end{gathered}$$

Hence, there is no any effect of magnetic field on the speed of the particle it means constant.

The acceleration can be calculated as

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{Fa}}{\mathrm{m}} \\ \mathrm{a}=\frac{\left|\mathrm{q}\right|\mathrm{vBsin}\left(\mathrm{\theta }\right)}{\mathrm{m}} \end{gathered}$$

Put all values than

$\mathrm{a}=\frac{\left(3.20\times {10}^{-19}\mathrm{C}\right)\left(35.6\times {10}^3\mathrm{m}/\mathrm{s}\right)\left(1.80\mathrm{T}\right)\sin \left(90°\right)}{\left(6.64\times {10}^{-27}\mathrm{kg}\right)}$

The magnetic field and velocity perpendicular to the acceleration so its radially inward and horizontal the curvature of the particles path.

Step 4: Determine why the speed of the particle does not change even though an unbalanced external force acts on it

The velocity is perpendicular the force. It means magnitude is not changed and direction of the velocity should be change. That’s why the speed of the particle does not change even though an unbalanced external force acts on it.