91Ó°ÊÓ

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

The force acting on it

${\stackrel{â‡\P Ä}{F}}_B=q\left(\stackrel{â‡\P Ä}{v_0}×\stackrel{â‡\P Ä}{B}\right)$

Velocity and magnetic field perpendicular to each other so

${\stackrel{â‡\P Ä}{F}}_B=\left|q\right|\left(\stackrel{â‡\P Ä}{v_0}×\stackrel{â‡\P Ä}{B}\right)$ ...... (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 $\stackrel{â‡\P Ä}{{\mathrm{F}}_{\mathrm{B}}}=\left|\mathrm{q}\right|\mathrm{vBsin}\left(\mathrm{θ}\right)$where $\mathrm{θ}=90°\mathrm{So}{\mathrm{F}}_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vB}$

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

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

Put all the values in

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

Diameter is twice of radius so

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

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

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

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

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

The acceleration can be calculated as

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

Put all values than

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

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

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.