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A proton of velocity $\left( {3\hat i + 2\hat j} \right)\,ms^{-1}$ enters a magnetic field of $(2\hat j + 3\hat k)\, tesla$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times10^8\,Ckg^{-1}$)
$2.88 \times {10^8}\left( {2\hat i - 3\hat j} \right)\,m/s^2$
$2.88 \times {10^8}\left( {2\hat i - 3\hat j + 2\hat k} \right)\,m/s^2$
$2.88 \times {10^8}\left( {2\hat i + 3\hat k} \right)\,m/s^2$
$2.88 \times {10^8}\left( {\hat i - 3\hat j + 2\hat k} \right)\,m/s^2$
Solution
Given ${\text{v}} = (3{\text{i}} + 2{\text{j}})\,{\text{m}}{{\text{s}}^{ – 1}}$ and $\overrightarrow {\text{B}} = (2{\text{j}} + 3{\text{k}})$ $tesla$.
Force experienced by the proton is
$\overrightarrow {\text{F}} = {\text{q}}(\overrightarrow {\text{v}} \times \overrightarrow {\text{B}} ) = {\text{q}}(3\widehat {\text{i}} + 2\widehat {\text{j}}) \times (2\widehat {\text{j}} + 3\widehat {\text{k}})$
$ = q(6\hat i – 9\hat j + 6\hat k)$
$ = 3{\text{q}}(2\widehat {\text{i}} – 3\widehat {\text{j}} + 2{\text{k}})\,{\text{N}}$
Acceleration $ = \frac{{\text{F}}}{{\text{m}}} = \frac{{3{\text{q}}}}{{\text{m}}}(2{\text{i}} – 3{\text{j}} + 2{\text{k}})$
$ = 3 \times \left( {0.96 \times {{10}^8}} \right)(2i – 3j + 2k)$
$ = 2.88 \times {10^8}(2\widehat {\text{i}} – 3\widehat {\text{j}} + 2{\text{k}}){\mkern 1mu} {\text{m}}{{\text{s}}^{ – 2}}$
Hence the correct choice is $( 2)$
