Question

A 150-V battery is connected across two parallel metal plates of area 28.5 cm² and separation 8.20 mm.A beam of alpha particles (charge +2e, mass 6.64*10^-27kg) is accelerated from rest throguh a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field.Whatmagnitude and direction of matnetic field are needed so that the alpha particles emerge undeflected from between the plates?

Answer 1

The voltage of the battery, \(V=150 \mathrm{~V}\)

The area of the metal plate is

$$ \begin{aligned} A &=28.5 \mathrm{~cm}^{2} \\ &=\left(28.5 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right) \\ &=28.5 \times 10^{-4} \mathrm{~m}^{2} \end{aligned} $$

The separation between two parallel metal plates is

$$ \begin{aligned} d &=8.20 \mathrm{~mm} \\ &=(8.20 \mathrm{~mm})\left(\frac{10^{-3} \mathrm{~m}}{1 \mathrm{~mm}}\right) \\ &=8.20 \times 10^{-3} \mathrm{~m} \end{aligned} $$

The mass of the alpha particle, \(m=6.64 \times 10^{-27} \mathrm{~kg}\)

The charge of the alpha particle, \(q=+2 e\)

The charge of the electron, \(e=1.60 \times 10^{-19} \mathrm{C}\)

The potential difference is

$$ \begin{aligned} \Delta V &=1.75 \mathrm{kV} \\ &=(1.75 \mathrm{kV})\left(\frac{10^{3} \mathrm{~V}}{1 \mathrm{kV}}\right) \\ &=1750 \mathrm{~V} \end{aligned} $$

The work done on the alpha particle is

$$ W=q \Delta V $$

Here, the potential difference is \(\Delta V\)

The charge of the alpha particle is \(q\)

The change of kinetic energy of the alpha particle is given by

$$ \Delta E=\frac{1}{2} m v^{2}-\frac{1}{2} m v_{0}^{2} $$

Here, the initial velocity of the alpha particle \(v_{0}=0 \mathrm{~m} / \mathrm{s}\)

The final velocity of the alpha particle is \(v\)

Therefore, the above equation becomes

$$ \Delta E=\frac{1}{2} m v^{2} $$

From the conservation of energy, the work done on the alpha particle is equal to the change of kinetic energy of the alpha particle. Such that

$$ \begin{aligned} \frac{1}{2} m v^{2} &=q \Delta V \\ v^{2} &=\frac{2 q \Delta V}{m} \\ v &=\sqrt{\frac{2(2 e) \Delta V}{m}} \\ &=\sqrt{\frac{4 e \Delta V}{m}} \quad \ldots \ldots .(1) \end{aligned} $$

Using equation (1), the velocity of the alpha particle is

$$ \begin{aligned} v &=\sqrt{\frac{4\left(1.60 \times 10^{-19} \mathrm{C}\right)(1750 \mathrm{~V})}{\left(6.64 \times 10^{-27} \mathrm{~kg}\right)}} \\ &=4.107 \times 10^{5} \mathrm{~m} / \mathrm{s} \end{aligned} $$

The electric field between two parallel metal plates due to battery is

$$ \begin{aligned} E &=\frac{V}{d} \\ &=\frac{(150 \mathrm{~V})}{\left(8.20 \times 10^{-3} \mathrm{~m}\right)} \\ &=18292.68 \mathrm{~V} / \mathrm{m} \end{aligned} $$

The electrostatic force is given by

$$ F_{E}=q E $$

The magnetic force is given by

$$ F_{B}=q v B $$

Here, the magnetic field strength is \(B\)

If the alpha particles are not deflected, then the magnetic force must be canceling the electric force. Therefore

$$ \begin{aligned} q v B &=q E \\ B &=\frac{E}{v} \quad \ldots \ldots (2)\end{aligned} $$

Using equation (2), the magnitude of the magnetic field is

$$ \begin{aligned} B &=\frac{(18292.68 \mathrm{~V} / \mathrm{m})}{\left(4.107 \times 10^{5} \mathrm{~m} / \mathrm{s}\right)} \\ &=0.0445 \mathrm{~T} \end{aligned} $$

According to given problem, the magnetic field \(\vec{B}\) is perpendicular to the electric field \(\vec{E}\).

If the particles are moving to the right, then the electric field is along upward direction.

From the right hand rule, the direction of the magnetic field is pointed to the out of the page.