Find the direction and magnitude of the minimum velocity of charge q = +2e that will produce the magnetic force F shown in the figure below. (Measure the angle counterclockwise from the x axis in the xz plane. Take the magnitudes of F as 1.96 x 10-11 N and the magnitude of B as 2.19 T.) magnitude   direction

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Find the direction and magnitude of the minimum velocity of charge q = +2e that will produce the magnetic force F shown in the figure below. (Measure the angle counterclockwise from the x axis in the xz plane. Take the magnitudes of F as 1.96 x 10-11 N and the magnitude of B as 2.19 T.)

magnitude  
direction
45.0°
Transcribed Image Text:45.0°
Expert Solution
Step 1

Given:

Force, F = 1.96×10-11 NMagnetic field, B = 2.19 TCharge, q = +2e = 2×1.6×10-19 C = +3.2×10-19 C

 

Solution:

We know that, the force F is perpendicular to the velocity 'v'  and magnetic force 'B'.

Let us consider velocity is perpendicular to magnetic field B and electric field E, therefore velocity 'v' is 45° counter-clockwise from x-axis in x-z plane and the direction of particle is along : x i^ - z j^

Consider the following equation:

 F = q v Bwhere, q is the charge             v is velocity and            B is magnetic field

Substituting the given values in above equation, we get:

F = q v B1.96×10-11 N = 3.2×10-19C×v×2.19 T

 

Re-arranging the above equation:

v = 1.96×10-11 N3.2×10-19×2.19 T= 2.8×107 m/s

 

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