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A charged particle projected in a magnet...

A charged particle projected in a magnetic field
`B=(3hati+4hatj)xx10^-2T`
The acceleration of the particle is found to be
`a=(xhati+2hatj)m/s^2`
find the value of `x`

Text Solution

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 ENGLISH-ELECTRODYNAMICS-Exercise-1 PART-1
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