Question

What is the minimum magnetic induction of a magnetic field to be applied at right angles to the electric field so that none of the electrons will reach the positive plate?

• A. $B=\sqrt{\frac{2Vm}{e{d}^2}}$
• B. $B=\sqrt{\frac{Vm}{e{d}^2}}$
• C. $B=\sqrt{\frac{2V}{me{d}^2}}$
• D. $B=\sqrt{\frac{V}{me{d}^2}}$

Answer 1

The correct option is A $B=\sqrt{\frac{2Vm}{e{d}^2}}$

$\text{Electric field}E=\frac{V}{d}$

Force applied on the electron due to electric field $F_E=q.E{m}_e.a=e.\frac{V}{d}d=\frac{eV}{m_{e}a}a=\frac{eV}{d{m}_e}$

$v^2=u^2+2a.s{v}^2=0^2+2\times \frac{eV}{d.m_e}\times dv=\sqrt{\frac{2eV}{m_e}}$

Force applied on the charge due to magnetic field $F=q.(\overrightarrow{v}\times \overrightarrow{B})\frac{m_e{v}^2}{r}=q\times vB\sin {90}^{\circ }r=\frac{m_e.v}{q.B}$

For not striking to the plate $d>rd>\frac{m_{e}V}{q.B}d>\frac{m_e\times \sqrt{\frac{2eV}{m_e}}}{e.B}d>\sqrt{\frac{2m_{e}V}{e.B^2}}B>\sqrt{\frac{2m_{e}V}{e.d^2}}$ (particle will start circular motion)

Hence option 'A' is correct.