Question

A charged particle is moving with the speed of 10 ^6 m/s at an angle 30 degree to an external magnetic field at B=10-4 J if q /m of the particle is 10 ^8 C/Kg find 1. Radius of trajectory 2. Pitch Please answer me fast please

Answer 1

A charge particle is moving with the speed of 10^6 m/s at an angle 30° to an external magnetic field, B = 10¯⁴ T. here q/m of the particle is 10^8 C/kg.

To find : (1) radius of the trajectory (2) pitch

solution : we know, radius of the trajectory $r=\frac{mv_{perp}}{qB}$

and pitch , $P=\frac{2\pi m v_{parallel}}{qB}$

here, B = 10¯⁴ T , q/m = 10^8 C/kg

$v_{perp}$ = 10^6 sin30° = 0.5 × 10^6 m/s

$v_{parallel}$ = 10^6 cos30° = 0.866 × 10^6 m/s

now r = (m/q) × ($v_{perp}$/B)

= 1/(q/m) × (0.5 × 10^6)/(10¯⁴)

= 10^-8 × 0.5 × 10^10

= 0.5 × 10² = 50 m

again, P = 2π (m/q) ($v_{parallel}$/B)

= 2 × 3.14 × 1/(q/m) × (0.866 × 10^6/10^-4)

= 6.28 × 10^-8 × 0.866 × 10^10

= 6.28 × 0.866 × 10² m

= 5.43848 × 10² m

= 543.848 ≈ 544 m

Therefore the radius of the trajectory is 50m and pitch of it is 544 m (approx)