Chapter 31, Problem 39P

(a)

To determine

The magnitude of the force, which is exerted on an electron in time varying magnetic field.

Answer to Problem 39P

The magnitude of the force which is exerted on an electron in time varying magnetic field is $8.01\times {10}^{-21}\text{N}$.

Explanation of Solution

Follow the right-handed cylindrical coordinate system.

Write the expression for Faraday’s law.

    ${\displaystyle \oint E\cdot dl}=-A\frac{dB\left(t\right)}{dt}$                                                                                                   (I)

Here, $E$ is the vortex electric field produced by time varying magnetic field, $dl$ is the line element, $A$ is the area, $t$ is the time and $B\left(t\right)$ is the magnetic field.

Write the expression for area.

    $A=\pi R^2$

Here, $A$ is area, $R$ is radius of circular region in which the magnetic field is varying.

Write the expression for length element.

    ${\displaystyle \oint dl}=2\pi r_1\widehat{\text{φ}}$

Here, $r_1$ is the radius of the circle under which the line integral is calculated and $\widehat{\text{φ}}$ is unit vector in anticlockwise direction, as seen from above.

Write the expression for magnitude of force on electron.

    $F=eE$                                                                                                                   (II)

Here, $F$ is force and $e$ is charge of electron.

Let the electric field be in clockwise $-\widehat{\text{φ}}$ direction then equation (I) takes the following form.

    $E{\displaystyle \oint \left(-\widehat{\text{φ}}\right)\cdot \left(dl\right)}=-A\frac{dB\left(t\right)}{dt}$

Substitute $\left(2\pi r_1\widehat{\text{φ}}\right)$ for ${\displaystyle \oint dl}$, $\pi R^2$ for $A$ and $\left(2.00t^3-4.00t^2+0.800\right)$ for $B\left(t\right)$ in above equation to find $E$.

    $\begin{array}{c}E\left(-\widehat{\text{φ}}\right)\cdot \left(2\pi r_1\widehat{\text{φ}}\right)=-\pi R^2\frac{d}{dt}\left(2.00t^3-4.00t^2+0.800\right)\\ E2\pi r_1\left(-\widehat{\text{φ}}\cdot \widehat{\text{φ}}\right)=-\pi R^2\left(6t^2-8t\right)\\ E\left(-\widehat{\text{φ}}\cdot \widehat{\text{φ}}\right)=-\frac{R^2}{2r_1}\left(6t^2-8t\right)\\ E=\frac{R^2}{2r_1}\left(6t^2-8t\right)\left(-\widehat{\text{φ}}\right)\end{array}$

Substitute $\frac{R^2}{2r_1}\left(6t^2-8t\right)\left(-\widehat{\text{φ}}\right)$ for $E$ in equation (II) to find force.

    $F=\frac{e{R}^2}{2r_1}\left(6t^2-8t\right)\left(-\widehat{\text{φ}}\right)$                                                          (III)

Conclusion:

Substitute $2.00\text{s}$ for $t$, $2.50\text{cm}$ for $R$, $1.6\times {10}^{-19}\text{C}$ for $e$ and $5\text{cm}$ for $r_1$ in equation (III) to find force.

    $\begin{array}{c}F=\frac{\left(1.6\times {10}^{-19}\text{C}\right){\left(2.50\text{cm}\right)}^2}{2\left(5\text{cm}\right)}\left(6{\left(2.00\text{s}\right)}^2-8\left(2.00\text{s}\right)\right)\left(-\widehat{\text{φ}}\right)\\ =\frac{\left(1.6\times {10}^{-19}\text{C}\right){\left(2.50\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{2\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\left(6{\left(2.00\text{s}\right)}^2-8\left(2.00\text{s}\right)\right)\left(-\widehat{\text{φ}}\right)\\ =\frac{\left(1.6\times {10}^{-19}\text{C}\right){\left(0.0250\text{m}\right)}^2}{2\left(0.05\text{cm}\right)}\left(8\right)\left(-\widehat{\text{φ}}\right)\\ =8.01\times {10}^{-21}\left(-\widehat{\text{φ}}\right)\text{N}\end{array}$

Therefore, the magnitude of the force on electron is $8.01\times {10}^{-21}\text{N}$.

(b)

To determine

The direction of the force, which is exerted on an electron in time varying magnetic field.

Answer to Problem 39P

The direction of force exerted on an electron is clockwise if seen from above.

Explanation of Solution

The direction of the electric field has to be in the clockwise direction. It is clear from equation (I) that the direction of $E$ must be $-\widehat{\text{φ}}$ to hold the sign convention on both side. Direction of force will be same as of vortex electric field. Therefore direction of force on electron is $-\widehat{\text{φ}}$ direction. $-\widehat{\text{φ}}$ denotes clockwise direction in right handed cylindrical coordinate system.

Therefore, direction of force exerted on the electron is clockwise if seen from above.

(c)

To determine

The time when force on the electron is equal to zero.

Answer to Problem 39P

The force on the electron is equal to zero at $t=0\text{s}$ or $t=1.33\text{s}$.

Explanation of Solution

In the time varying magnetic field, force on the electron must be zero for particular instant of time.

Conclusion:

Substitute $0$ for $F$ in equation (III) to find time.

    $\begin{array}{c}\frac{e{R}^2}{2r_1}\left(6t^2-8t\right)\left(-\widehat{\text{φ}}\right)=0\\ \left(6t^2-8t\right)=0\\ 2t\left(3t-4\right)=0\\ t=0\text{s}\end{array}$

Further solve the equation.

    $\begin{array}{c}\left(3t-4\right)=0\\ t=\frac{3}{4}\text{s}\\ =1.\text{33}\text{s}\end{array}$

Therefore, force on electron is equal to zero at $t=0\text{s}$ or $t=1.33\text{s}$.