Answer:
The speed of the particle is [tex]1.807\times 10^{7}\ m/s[/tex].
Explanation:
Given:
Mass of the alpha-particle (m) = 6.64 × 10⁻²⁷ kg
Charge of the particle (q) = 3.2 × 10⁻¹⁹ C
Magnitude of magnetic field (B) = 1.5 T
Radius of the circular path (r) = 25 cm = 0.25 m [1 cm = 0.01 m]
Speed of the particle (v) = ?
We know that, when a charged particle enters a uniform magnetic field, the path traced by the charged particle is circular and the radius of the circular path is given by the formula:
[tex]r=\dfrac{mv}{qB}[/tex]
Expressing the above formula in terms of 'v', we get:
[tex]v=\dfrac{qBr}{m}[/tex]
Now, plug in the values given and solve for 'v'. This gives,
[tex]v=\dfrac{3.2\times 10^{-19}\ C\times 1.5\ T\times 0.25\ m}{6.64\times 10^{-27}\ kg}\\\\\\v=\dfrac{1.2\times 10^{-19}}{6.64\times 10^{-27}}\ m/s\\\\v=1.807\times 10^{7}\ m/s[/tex]
Therefore, the speed of the particle is [tex]1.807\times 10^{7}\ m/s[/tex].
