A charged particle carrying charge `q=1muc` moves in uniform magnetic with velocity `v_1=10^6m//s` at angle `45^@` with `x`-axis in the `xy`-plane and experiences a force `F_1=5sqrt2N` along the negative `z`-axis. When te same particle moves with velocity `v_2=10^6m//s` along the `z`-axis it experiences a force `F_2` in `y`-direction. Find
a. the magnitude and direction of the magnetic field
b. the magnitude of the force `F_2`.

Explanation

To solve the problem, we will break it down into two parts: finding the magnitude and direction of the magnetic field (part a), and then calculating the magnitude of the force \( F_2 \) (part b).

### Part a: Finding the Magnitude and Direction of the Magnetic Field

1. **Given Data:**
- Charge of the particle, \( q = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Velocity \( v_1 = 10^6 \, m/s \) at an angle of \( 45^\circ \) with the x-axis in the xy-plane.
- Force \( F_1 = 5\sqrt{2} \, N \) along the negative z-axis.
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