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A particle with mass1.81×10-3kgand a charge of has1.22×10-8C, at a given instant, a velocityV=(3.00×104m/s).What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic fieldB=(1.63T)i+(0.980T)j^?

Short Answer

Expert verified

The acceleration isa=-(0.330m/s2)k^

Step by step solution

01

Important Concepts

The magnetic field force is given by

FB=q(v×b)F=qvBsinθ

Where q is the charge of the particle, V is the velocity and B is the magnetic field

02

Determination of magnitude and direction of the particles

By the second law of motion we know that the force on a body is

F=ma

And the magnetic field force is given by

F=q(v×B)

Equating the two we get

ma=qv×Ba=qv×BmSubstituteallthevalueintheaboveequationa=1.22×10-8C3.0×104m/sj×1.63ti+0.980Tj1.81×10-3kga=-(0.330m/s3)k^Hence,theaccelerationisa=-(0.330m/s3)k^.

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