A 2.5 pC charge is initially moving at a speed of 200 m/s at an angle of 43° N of E in a uniform magnetic field directed directly up (N). a) Sketch the situation. b) Find the magnitude of the force exerted on the charge. c) Find the direction of the force.
Added by Guillermo F.
Step 1
Given: Q = 2.5 pC = 2.5 x 10^-12 C v = 200 m/s B = given as directed directly up (N), so B = 1 T (assuming a magnetic field strength of 1 Tesla) θ = 43° Plugging in the values: F = (2.5 x 10^-12 C)(200 m/s)(1 T)sin(43°) F = 2.5 x 10^-12 x 200 x sin(43°) F = 2.5 Show more…
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