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If a particle of charge ${10^{ - 12}}\,coulomb$ moving along the $\hat x - $ direction with a velocity ${10^5}\,m/s$ experiences a force of ${10^{ - 10}}\,newton$ in $\hat y - $ direction due to magnetic field, then the minimum magnetic field is
$6.25 \times {10^3}\,tesla$ in $\hat z - $ direction
${10^{ - 15}}\,tesla$ in $\hat z - $ direction
$6.25 \times {10^{ - 3}}\,tesla$ in $\hat z - $ direction
${10^{ - 3}}\,tesla$ in $\hat z - $ direction
Solution
(d) $F = qvB\sin \theta $ $==>$ $B = \frac{F}{{qv\sin \theta }}$
${B_{\min }} = \frac{F}{{qv}}$(when $\theta$ = $90^o$)
$\therefore \;{B_{\min }} = \frac{F}{{qv}} = \frac{{{{10}^{ – 10}}}}{{{{10}^{ – 12}} \times {{10}^5}}} = {10^{ – 3}}$ $Tesla$ in $U = \frac{{{B^2}}}{{2{\mu _0}}},$ $\hat z – $ direction.
