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Byju's Answer

Standard XII

Physics

Force on a Current Element

A charged par...

Question

A charged particle having kinetic energy E enters normally a region of uniform magnetic field between two plates $p_{1}$ and $p_{2}$ as shown in the figure. If the particle just misses hitting the plate $p_{2}$ then the magnetic field B in the region between the plates is :

\frac{m E}{2 q d}

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\frac{2 m E}{q d}

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\frac{\sqrt{m E}}{2 q d}

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\frac{\sqrt{2 m E}}{q d}

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Solution

The correct option is D $\frac{\sqrt{2 m E}}{q d}$
Kinetic energy of a particle is given by,
$\frac{1}{2} m v^{2} = E$
V $= \sqrt{\frac{2 E}{m}}$
And $r = d$
$\frac{m v}{q B} = d$
$B = \frac{m}{q d} \sqrt{\frac{2 E}{m}}$
$B = \frac{\sqrt{2 m E}}{q d}$

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A charged particle having kinetic energy E enters normally a region of uniform magnetic field between two plates p1 and p2 as shown in the figure. If the particle just misses hitting the plate p2 then the magnetic field B in the region between the plates is :

The correct option is D √2mEqdKinetic energy of a particle is given by,12mv2=EV=√2EmAnd r=dmvqB=dB=mqd√2EmB=√2mEqd

A charged particle having kinetic energy E enters normally a region of uniform magnetic field between two plates $p_{1}$ and $p_{2}$ as shown in the figure. If the particle just misses hitting the plate $p_{2}$ then the magnetic field B in the region between the plates is :

The correct option is D $\frac{\sqrt{2 m E}}{q d}$
Kinetic energy of a particle is given by,
$\frac{1}{2} m v^{2} = E$
V $= \sqrt{\frac{2 E}{m}}$
And $r = d$
$\frac{m v}{q B} = d$
$B = \frac{m}{q d} \sqrt{\frac{2 E}{m}}$
$B = \frac{\sqrt{2 m E}}{q d}$