Question

An alpha-particle (m = 6.64 × 10−27 kg, q = 3.2 × 10−19 C) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. What is the speed of the particle?

Answer 1

The speed of the particle is
`1.807* 10^(7)\ m/s`.

Step-by-step 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:

`r=(mv)/(qB)`

Expressing the above formula in terms of 'v', we get:

`v=(qBr)/(m)`

Now, plug in the values given and solve for 'v'. This gives,

`v=(3.2* 10^(-19)\ C* 1.5\ T* 0.25\ m)/(6.64* 10^(-27)\ kg)\\\\\\v=(1.2* 10^(-19))/(6.64* 10^(-27))\ m/s\\\\v=1.807* 10^(7)\ m/s`

Therefore, the speed of the particle is
`1.807* 10^(7)\ m/s`.