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An experimenter's diary reads as follows...

An experimenter's diary reads as follows, "a charged particle is projected in a magnetic field of `(7.0hati-3.0hatj)xx10^(-3)T`.The acceleration of the particle is found to be `(xhati+7.0hatj)xx10^(-6)m//s^(2)`.Find the value of `x`.

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Verified by Experts

The correct Answer is:
C
`vecF=q(vecvxxvecB)`
`vecF_|_vecBrArrveca is _|_"to" vecB`
So,`veca.vecB=0`
`7x-21=0`
`x=3.0`
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RESONANCE-ELECTRODYNAMICS-Exercise-1 PART-1
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