PSS 27.1: Magnetic Forces 4 of Part C Constants You can check your result by com

Question

PSS 27.1: Magnetic Forces 4 of Part C Constants You can check your result by comparing its magnitude to the magnitude the acceleration would have if the particle's velocity had the s magnitude but it was perpendicular to the magnetic field. Find the value of the expression quB/m (the magnitude of a when v is perpendicular to B), where q is the magnitude of the charge the magnitude of the velocity, B is the magnitude of the magnetic field, and m is the mass of the particle. Express your answer in meters per second squared. Learning Goal: To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81x10-3 kg and a charge of 1.22x10-8 C has, at a given instant, a velocity v (3.00 x 101 m/s)j. What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field B = (1.63 T)i + (0.980 T) View Available Hint(s) /? Submit

Explanation / Answer

The problem is that v is not completely perpendicular to B. v has a 'j' component and B has an 'i and j components. If v was completely perpendicular, there would be no j component. Since they are not perpendicular, you have to use the real definition of magnetic force...

F = m*a = q*(v cross B)

a = q/m * (v cross B)

v cross B = (0i + 30000j m/s) cross (1.63i + 0.980j T) = 0 i + 0 j - 48900 k m/s*T

a = 1.22 * 10^-8 C / 1.81 * 10^-3 kg * (0i + 0j - 48900 k m/s*T) = 0i + 0j - 0.3296 k m/s^2

Magnitude = 0.3296 m/s^2

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