Answer:
The magnitude of the magnetic field is 4.73 T.
Explanation:
Given that,
Charge of the particle, [tex]q=-4\ \mu C=-4\times 10^{-6}\ C[/tex]
It is moving with a velocity of 2 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s². We need to find the magnitude of magnetic field.
When it enters in magnetic field,
[tex]qvB\sin \theta=ma[/tex]
B is magnetic field
[tex]B=\dfrac{ma}{qv\sin \theta}\\\\B=\dfrac{5\times 10^{-3}\times 5.8}{4\times 10^{-6}\times 2\times 10^3\sin (50)}\\\\B=4.73\ T[/tex]
So, the magnitude of the magnetic field is 4.73 T.
