An electron moves in the X-Y plane with a speed of `10^(6)` m/s. Its velocity vector `vec v` makes an angle `60^(@)` with X axis. There is a magnetic field of magnitude `10^(-2)` `Wb/m^(2)` directed along the Y axis. Calculate the magnetic force vector acting on the electron. (electron charge = `– 1.6 xx 10^(-19) columbs)`

Explanation

To calculate the magnetic force vector acting on the electron, we will follow these steps:

### Step 1: Determine the components of the velocity vector
Given that the speed of the electron is \( v = 10^6 \, \text{m/s} \) and it makes an angle of \( 60^\circ \) with the X-axis, we can find the components of the velocity vector \( \vec{v} \).

\[ v_x = v \cos(60^\circ) = 10^6 \cdot \frac{1}{2} = 5 \times 10^5 \, \text{m/s} \]
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