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 isBe-pp-BeBeppe-B

Question

Toppr Admin · Accepted Answer

Answer is C-Force in a magnetic fieldF-qVBand F-mV2r -xA0- -By Newton-apos-s Law-so- mV2r-qVBmV - qrB-x2192-r-mVqB-pBe -xA0- -xA0-x2235-mV-p-xA0-xA0-and-xA0-xA0-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
Be/p
p/Be
Bep
pe/B

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

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

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

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 isBe/pp/BeBeppe/B

Answer 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
Be/p
p/Be
Bep
pe/B

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

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

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