A charged particle of mass m and charge q is accelerated through a potential difference of V volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by

Question

A charged particle of mass m and charge q is accelerated through a potential difference of V volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by (A) √(V m/q B2) (B) (2 V m/q B2) (C) √(2 V m/q) ⋅((1/B)) (D) √(V m/q) ⋅((1/B))

Tardigrade Admin · Accepted Answer

The speed of particle just before entering into magnetic field is v0. ∴ q V=(1/2) m v02 ⇒ v0=√((2 q V/m)) ∴ r=(m v0/q B)=(m/q B) √((2 q V/m)) =(1/B) √((2 m V/q))

Q. A charged particle of mass $m$ and charge $q$ is accelerated through a potential difference of $V$ volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by

$\frac{Vm}{q B ^{2}}$

$\frac{2 Vm}{q B ^{2}}$

$\frac{2 Vm}{q} \cdot (\frac{1}{B})$

$\frac{Vm}{q} \cdot (\frac{1}{B})$

The speed of particle just before entering into magnetic field is $v_{0}$ .
$∴ q V = \frac{1}{2} m v_{0}^{2}$
$\Rightarrow v_{0} = (\frac{2 q V}{m})$
$∴ r = \frac{m v _{0}}{qB} = \frac{m}{qB} (\frac{2 q V}{m})$
$= \frac{1}{B} (\frac{2 mV}{q})$

Q. A charged particle of mass m and charge q is accelerated through a potential difference of V volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by

qB2Vm​​

qB22Vm​

q2Vm​​⋅(B1​)

qVm​​⋅(B1​)