Answer:
Magnetic field, 0.046 T
Explanation:
Given that,
Mass of the charged particle, m = 5 g
Charge, [tex]q=-70\ \mu C=-70\times 10^{-6}\ C[/tex]
Electric field, E = 700 N/C
Speed of the charge particle, v = 30 km/s = 30000 m/s
To find,
The magnitude of the magnetic field in this region.
Solution,
The charged particle is moving horizontally with a constant speed, the gravitational force is balanced by the magnetic force and the electric force. [tex]mg=qvB+qE[/tex]
[tex]mg-qE=qvB[/tex]
[tex]B=\dfrac{mg-qE}{qv}[/tex]
[tex]B=\dfrac{5\times 10^{-3}\times 9.8-(-70\times 10^{-6})\times 700}{70\times 10^{-6}\times 30000}[/tex]
B = 0.046 T
Therefore, the magnitude of the magnetic field in this region is 0.046 T. Hence, this is the required solution.
