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Chapter 4: Q4E (page 912)

A particle with mass1.81脳10-3kgand a charge of has1.22脳10-8C, at a given instant, a velocityV鈬赌=(3.00脳104m/s).What are the magnitude and direction of the particle鈥檚 acceleration produced by a uniform magnetic fieldB=(1.63T)i+(0.980T)j^?

Short Answer

Expert verified

The acceleration isa鈬赌=-(0.330m/s2)k^

Step by step solution

01

Important Concepts

The magnetic field force is given by

FB=q(v脳b)F=qvBsin胃

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鈬赌=ma鈬赌

And the magnetic field force is given by

F=q(v脳B)

Equating the two we get

ma鈬赌=qv脳Ba鈬赌=qv脳BmSubstituteallthevalueintheaboveequationa鈬赌=1.22脳10-8C3.0脳104m/sj脳1.63ti鈬赌+0.980Tj鈬赌1.81脳10-3kga鈬赌=-(0.330m/s3)k^Hence,theaccelerationisa鈬赌=-(0.330m/s3)k^.

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Fig. E25.29.

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