A particle with mass 1.8110 ?3 kg and a charge of 1.2210 ?8 C has, at a given in

Question

A particle with mass 1.8110?3kg and a charge of 1.2210?8C has, at a given instant, a velocity v? =(3.00104m/s)j^. What are the magnitude and direction of the particles acceleration produced by a uniform magnetic field B? =(1.63T)i^+(0.980T)j^?

A)Find the acceleration vector for the charge.

Enter the x, y, and z components of the acceleration in meters per second squared separated by commas.

B)

You can check your result by comparing its magnitude to the magnitude the acceleration would have if the particle's velocity had the same magnitude but it was perpendicular to the magnetic field.

Find the value of the expression qvB/m (the magnitude of a?  when v?  is perpendicular to B? ), where qis the magnitude of the charge, v is the magnitude of the velocity, B is the magnitude of the magnetic field, and m is the mass of the particle.

Can you please explain how to find these values? Thanks

Explanation / Answer

0.385 is the answer for part B

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