Question

an alpha particle is travelling at right angles to magnetic field with a speed of 6 *10^5 m/s.the strength of magnetic field is 0.2T . calculate acceleration of alpha particle

Answer 1

3*10^6 m/s^2
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Answer 2

For the given question we have to find force experienced by alpha particle, so formula to be used will be:

                                          F = qvB sinθ

We have values for B= 0.2 T,

$v=6\times 10^{ 5 }m/s, \theta=90°$

Now alpha particle’s charge = 2 × charge on proton

                                   $=2\times (1.6\times 10^{-19}) C$

                                   $= 3.2\times 10^{-19} C$

Now applying the values in formula,

                                   $F= (3.2\times 10^{-19})\times (6\times 10^5)\times (0.2)\times sin90degree$

                                   $=3.84\times 10^{-14} N$

We also have to find acceleration

                                          = F/m

                                    $= \frac {3.84\times 10^{-14}} {6.44\times 10^{-27}}$

                                    $=5.96\times 10^{12}$ is our answer.