Question

An electron is accelerated from rest through potential difference Vand then enters a region of uniform magnetic field, where itundergoes uniform circular motion. Figure 28-38 gives the radius rof thatmotion versus V1/2. The vertical axis scale is set by${\mathbf{r}}_s\mathbf{=}\text{3.0mm}$and the horizontal axis scale is set by ${\mathbf{V}}_s^{\frac{1}{2}}\mathbf{=}\text{40.0}\mathbf{}{\mathbf{V}}^{\frac{1}{2}}$What is the magnitude of the magnetic field?

Short answer

Magnitude of magnetic field is $B=6.7\times {10}^{-2}\text{T}$

Step-by-step solution

Given

i) Scale on y axis is $r_s=3.\text{0mm}$

ii) Scale on x axis is $V_s^{\frac{1}{2}}=40.0V^{\frac{1}{2}}$

iii) Figure 28-38 is the graph of radius vs. potential.

Determining the concept

Use the concept of conservation of mechanical energy and magnetic and centripetal force. Rearrangetheenergy conservation equation for velocity. Also from the centripetal force and magnetic force equation, writetheequation for velocity. Equating these two velocities’ equations and plugging the value of slope from the graph$\frac{r}{{\left(\mathbf{V}\right)}^{\frac{1}{2}}}$find the magnetic field.

Formulae are as follow:

$P.E_i+K.E_i=P.E_f+K.E_f$

$F_B=qvB$

$F_{cp}=\frac{m{v}^2}{r}$

Here, F_B is magnetic force, B is magnetic field, v is velocity, m is mass,q is charge on particle, P.E is potential energy, K.E is kinetic energy,$F_{CP}$is centripetal force.

Determining the magnitude of magnetic field

Consider the expression as:

$P.E_i+K.E_i=P.E_f+K.E_f$

Here potential energy is the electrical potential energy.

$q{V}_i+\frac{1}{2}m{v}_i^2=q{V}_f+\frac{1}{2}m{v}_f^2$

Initially kinetic energy is zero, and finally potential energy is zero,

$q{V}_i=\frac{1}{2}m{v}_f^2$

Rewrite the equation as:

$v_f=\sqrt{\frac{2q{V}_i}{m}}$ …. (1)

The magnetic force provides the centripetal force, so:

$F_{cp}=F_B$

$\frac{m{v}_f^2}{r}=q{v}_{f}B$

Rewrite the equation as:

$v_f=\frac{qBr}{m}$

From equation (1) and (2):

$\frac{qBr}{m}={\left(\frac{2q{V}_i}{m}\right)}^{\frac{1}{2}}$

$\frac{qB}{m}\left(\frac{r}{V_i^{\frac{1}{2}}}\right)={\left(\frac{2q}{m}\right)}^{\frac{1}{2}}$

Determine the slope $\frac{r}{V_i^{\frac{1}{2}}}$from the graph.

In the graph for $V_s^{\frac{1}{2}}$, the value of $r=2.\text{0mm}$

$\frac{r}{V_i^{\frac{1}{2}}}=\frac{(2.0\times {10}^{-3})}{40.0V^{\frac{1}{2}}}$

Substitute the values in theabove equation:

$\frac{qB}{m}\left(\frac{(2.0\times {10}^{-3})}{40.0}\right)={\left(\frac{2q}{m}\right)}^{\frac{1}{2}}$

Rearranging this equation for B,

$B={\left(\frac{2q}{m}\right)}^{\frac{1}{2}}\left(\frac{40.0}{(2.0\times {10}^{-3})}\right)\left(\frac{m}{q}\right)$

Substitute the values of mass and charge of electron.

$B={\left(\frac{2(1.6\times {10}^{-19})}{(9.1\times {10}^{-31})}\right)}^{\frac{1}{2}}\left(\frac{40.0}{(2.0\times {10}^{-3})}\right)\left(\frac{9.1\times {10}^{-31}}{1.6\times {10}^{-19}}\right)$

$B={(0.3516\times {10}^{12})}^{\frac{1}{2}}(20\times {10}^3)(5.6875\times {10}^{-12})$

$B=67.44\times {10}^{-3}$

$B=6.7\times {10}^{-2}\text{\hspace{0.17em}T}$

Hence,the magnitude of magnetic field is $B=6.7\times {10}^{-2}\text{T}$

Therefore, use the concept of conservation of mechanical energy and magnetic and centripetal force. Using equations, rearrange for magnetic field. Plugging the values and value of slope from the graph, find the magnetic field.