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"(i)"(i) The force due to a magnetic field can be calculated using the formula F = q v B sin(theta)F=qvBsin(θ); where FF is the force, qq is the charge of the particle, vv is the particle's velocity, BB is the magnetic field, and thetaθ is the angle between the velocity vector and magnetic field vector:
Rightarrow F = 3.2 times 10^(- 19) times 550 times 0.045 times sin(52^(circ))⇒F=3.2×10−19×550×0.045×sin(52∘)
Rightarrow F = 3.2 times 10^(- 19) times 24.75 times 0.7880107536⇒F=3.2×10−19×24.75×0.7880107536
Rightarrow F = 3.2 times 10^(- 19) times 19.503266152⇒F=3.2×10−19×19.503266152
Rightarrow F = 6.241045169 times 10^(- 18)⇒F=6.241045169×10−18
therefore F = 6.24 times 10^(- 18)
Therefore, the force acting on the particle due to the field is around 6.24 times 10^(- 18) "N".
"(ii)" Now, we must find the acceleration of the particle due to the magnetic force acting on it.
Let's use the formula F = ma:
Rightarrow = 6.24 times 10^(- 18) = 6.6 times 10^(- 27) times a
Rightarrow 945,454,545.45 = a
therefore a = 9.46 times 10^(8)
Therefore, the acceleration on the particle due to vec(F_(B)) is around 9.46 times 10^(8) "m s"^(- 2).
"(iii)" As the particle accelerates, it increases in speed.
