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Standard XII

Physics

Velocity Selector

A particle ca...

Question

A particle carrying a charge $e$ perpendicular to a uniform magnetic field of induction B with a momentum p, then the radius of the circular path is

Be/p

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pe/B

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p/Be

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Bep

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Solution

The correct option is B p/Be
Answer is C.
Force in a magnetic field
$F = q V B$

and $F = \frac{m V^{2}}{r}$ (By Newton's Law)

so, $\frac{m V^{2}}{r} = q V B$

mV = qrB

$\to r = \frac{m V}{q B}$

$= \frac{p}{B e}$ ( $∵ m V = p a n d q = e$ )

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A particle carrying a charge e perpendicular to a uniform magnetic field of induction B with a momentum p, then the radius of the circular path is

The correct option is B p/BeAnswer is C.Force in a magnetic fieldF=qVBand F=mV2r (By Newton's Law)so, mV2r=qVBmV = qrB→r=mVqB=pBe (∵mV=p and q=e)

A particle carrying a charge $e$ perpendicular to a uniform magnetic field of induction B with a momentum p, then the radius of the circular path is

The correct option is B p/Be
Answer is C.
Force in a magnetic field
$F = q V B$

and $F = \frac{m V^{2}}{r}$ (By Newton's Law)

so, $\frac{m V^{2}}{r} = q V B$

mV = qrB

$\to r = \frac{m V}{q B}$

$= \frac{p}{B e}$ ( $∵ m V = p a n d q = e$ )