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A point charge Q =−40 nC is moving with the velocity of 6×10⁶m/s in the direction of (−0.48āx−0.6āy+0.64āz). What will be the force exerted on the moving charge by the fields, B = (2āx−3āy+5āz) mT and E = (2āx−3āy+5āz) kV/m?
(179āx+1003āy−833āz) N
(1.79āx+10.03āy−8.33āz) μN
(1.79āx+10.03āy−8.33āz) N
(179āx+1003āy−833āz) μN
(179āx+1003āy−833āz) μN
Practice and solve "A point charge Q =−40 nC is moving with the velocity of 6×106m/s in the direction of (−0.48āx−..." for Electrical - Electromagnetics Field Theory . The correct answer is Option D: (179āx+1003āy−833āz) μN. Detailed step-by-step solution, conceptual clarity, and formulas on Examoogle.
Option **(d) **