If a particle of charge {10^{ - 12}},coulomb | vedclass

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Question

(d) $F = qvB\sin 	heta $ $==>$ $B = \frac{F}{{qv\sin 	heta }}$
${B_{\min }} = \frac{F}{{qv}}$(when $	heta$ = $90^o$)
$	herefore \;{B_{\min }}

Vedclass · Accepted Answer

${10^{ - 3}}\,tesla$ in $\hat z - $ direction

Vedclass · Answer

(d) $F = qvB\sin 	heta $ $==>$ $B = \frac{F}{{qv\sin 	heta }}$
${B_{\min }} = \frac{F}{{qv}}$(when $	heta$ = $90^o$)
$	herefore \;{B_{\min }} = \frac{F}{{qv}} = \frac{{{{10}^{ - 10}}}}{{{{10}^{ - 12}} 	imes {{10}^5}}} = {10^{ - 3}}$ $Tesla$ in $U = \frac{{{B^2}}}{{2{\mu _0}}},$  $\hat z - $ direction.

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

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