An electron whose mass in 1×10^-31 kg and charge 1.6×10^-19, moving with speed of 3×10^6 parallel to axis ,enters acting in the magnetic field of 0.3 weber/m² parallel to z-axis. Find the force acting on it, acceleration and direction of acceleration.
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SOLUTION:-
Given:
Charge = 1.6× 10^-19 Coulomb
Velocity= 3× 10^6 m/sec,
B= 0.3 weber/m²
F= ? & ∅= 90°
From formula of Lorentz force, F=qvsin∅.
Force on electron,
Force on electron,
& the direction of force
Again,
Therefore,
From formula F= ma
Induced acceleration= F/m
Hope it helps ☺️
Answered by
0
Answer:
Given:
Charge = 1.6× 10^-19 Coulomb
Velocity= 3× 10^6 m/sec,
B= 0.3 weber/m²
F= ? & ∅= 90°
From formula of Lorentz force, F=qvsin∅.
Force on electron,
Force on electron,
& the direction of force
Again,
Therefore,
From formula F= ma
Induced acceleration= F/m
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