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An alpha particle is projected vertically upward with a speed of $3.0 \times 10^{4} km / s$ in a region where a magnetic field of magnitude $1.0 T$ exists in the direction south to north. The magnetic force that acts on the alpha – particle is [[1]] $\times 10^{- 12} N$ .

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Detailed Solution

An alpha particle is projected vertically upward with a speed of

3.0 \times 10^{4} km / s

in a region where a magnetic field of magnitude

1.0 T

exists in the direction south to north. The magnetic force that acts on the alpha – particle is

9.6 \times 10^{- 12} N

.
Given,
Magnetic field,

B = 1 T

Speed of alpha particle,

v = 3.0 \times 10^{4} km / s

Converting it to m/s,

v = 3.0 \times 10^{4} \times 10^{3} m / s

\Rightarrow v = 3.0 \times 10^{7} km / s

Also, we know that charge on alpha particle,

q = 2 \times charge on electron

Now, charge on the electron,

e = 1.6 \times 10^{- 19} C

So, charge on alpha particle,

q = 2 \times 1.6 \times 10^{- 19} C

\Rightarrow q = 3.2 \times 10^{- 19} C

The force experienced by moving charges in a magnetic field is given by

F = q (v \times B)

Putting the respective values in the above equation, we get

\Rightarrow F = 3.2 \times 10^{- 19} \times 3 \times 10^{7} \times 1

\Rightarrow F = 9.6 \times 10^{- 12} N

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