A charge q is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. The potential difference required to move it in a circle of radius 2r is

Question

A charge q is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. The potential difference required to move it in a circle of radius 2r is (A) 2 V . (B) 4 V (C) 1 V (D) 3 V

Tardigrade Admin · Accepted Answer

Radius of circular path: r = (√2mqV/qB) r propto√V where B is constant or V propto r2 (V2/V1) = ((r2/r1))2 (V2/V)= ((2r/r))2 4 ⇒ V2 = 4 V

Q. A charge q is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. The potential difference required to move it in a circle of radius 2r is

2 V .

4 V

1 V

3 V

Radius of circular path: $r = \frac{2 m q V}{qB}$
$r \propto V$ where B is constant
or $V \propto r^{2}$
$\frac{V _{2}}{V _{1}} = (\frac{r _{2}}{r _{1}})^{2}$
$\frac{V _{2}}{V} = (\frac{2 r}{r})^{2} 4$
$\Rightarrow V_{2} = 4 V$

Q. A charge q is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. The potential difference required to move it in a circle of radius 2r is

2 V .

4 V

1 V

3 V

Radius of circular path: r=qB2mqV​​ r∝V​ where B is constant or V∝r2 V1​V2​​=(r1​r2​​)2 VV2​​=(r2r​)24 ⇒V2​=4V

Radius of circular path: $r = \frac{2 m q V}{qB}$
$r \propto V$ where B is constant
or $V \propto r^{2}$
$\frac{V _{2}}{V _{1}} = (\frac{r _{2}}{r _{1}})^{2}$
$\frac{V _{2}}{V} = (\frac{2 r}{r})^{2} 4$
$\Rightarrow V_{2} = 4 V$