Answer :
Answer:
The magnitude of the magnetic field is [tex]9.3\times 10^{-5}\ T[/tex].
Explanation:
Given that,
Charge, [tex]q=-8.5\ \mu C=-8.5\times 10^{-6}\ C[/tex]
Speed of the charged particle, [tex]v=9\times 10^6\ m/s[/tex]
The angle between the velocity of the charge and the field is 56°.
The magnitude of force, [tex]F=5.9\times 10^{-3}\ N[/tex]
We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :
[tex]F=qvB\ \sin\theta[/tex]
B is the magnetic field.
[tex]B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T[/tex]
So, the magnitude of the magnetic field is [tex]9.3\times 10^{-5}\ T[/tex]. Hence, this is the required solution.