hep T Contact Us Cuthell, Physics, 9e College Physics I and II (PHY 2053 & Pract
ID: 1326470 • Letter: H
Question
hep T Contact Us Cuthell, Physics, 9e College Physics I and II (PHY 2053 & Practice Assignment Gradebook Assignment FULL SCREEN PRINTER VERSION ‘BACK RCES Chapter 21, Problem 07 em 19 m38 c em 53 A magnetic field has a magnitude of 0.0017 T, and an electric field has a magnitude of 5.6 × 103 NC. Both fields point in the same charge moves at a speed of 2.8 x 10% m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net a magnitude of 0.0017 T, and an electric field has a magnitude of 5.6 x 10 N/C. Both fields point in the same direction. A positive 2.2-pc of 2.8 x 106 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge. em 60 blenz 90 y Study Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work Question Attempts: 0 of 5 used SAVE POR LATER SUBMET ANSWERExplanation / Answer
We know that force acting on any charge of magnitude q moving with velocity v inside the magnetic field B is given by
F=q(v X B)
If both electric field E and magnetic field B are present. both field exert a force on the particle and the total force on the particle is equal to the vector sum of the electric field and magnetic field force.
F=qE+q(v X B)
=> F =q*(E) j + q*v*B k ( direction of E, B vector is taken as j, and plane out of the paper is k)
=> putting Values
F = (2.2*10^-6)*(5.6*103) j + (2.2*10^-6*2.8*10^6*.0017) k
F = 0.01232 j + 0.010472 k
Magnitude of F = 0.000261 N
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