An electron of mass textm and electric charge e is accelerated by a potential difference of V and then it enters in a magnetic field B perpendicular to the magnetic field lines. Radius of the circular path formed by electron is

Question

An electron of mass textm and electric charge e is accelerated by a potential difference of V and then it enters in a magnetic field B perpendicular to the magnetic field lines. Radius of the circular path formed by electron is (A) √2 eV / m (B) √2 Vm / eB2 (C) √2 Vm / eB (D) √2 Vm / e2 B

Tardigrade Admin · Accepted Answer

Bev=(mv2/r) or r=(mv/Be) as mv=√2 mT (.T = KE.) . So r=(√2 mT/Be) As the electron has been accelerated from rest through a potential difference of V volt, then T=eV ∴ r=√(2 mVe/B2 e2)=√(2 mV/B2 e)

Q. An electron of mass $m$ and electric charge $e$ is accelerated by a potential difference of $V$ and then it enters in a magnetic field $B$ perpendicular to the magnetic field lines. Radius of the circular path formed by electron is

$2 e V / m$

$2 Vm / e B^{2}$

$2 Vm / e B$

$2 Vm / e^{2} B$

$B e v = \frac{m v ^{2}}{r} or r = \frac{m v}{B e}$ as $m v = 2 m T$ $(T = K E)$ . So $r = \frac{2 m T}{B e}$
As the electron has been accelerated from rest through a potential difference of $V$ volt, then $T = e V$
$∴ r = \frac{2 mV e}{B ^{2} e ^{2}} = \frac{2 mV}{B ^{2} e}$

Q. An electron of mass m and electric charge e is accelerated by a potential difference of V and then it enters in a magnetic field B perpendicular to the magnetic field lines. Radius of the circular path formed by electron is

2eV/m​

2Vm/eB2​

2Vm/eB​

2Vm/e2B​