Question

The magnetic field in FIGURE EX30.19 is decreasing at the rate $0.10\text{T}/\text{s}$. What is the acceleration (magnitude and direction) of a proton initially at rest at points a to d?

Short answer

Part (a). the magnitude acceleration of the proton is$4.8\times {10}^4\text{m}/{\text{s}}^2$

Part (b). the magnitude acceleration of the proton is $0\text{m}/{\text{s}}^2$

Part (c). the magnitude acceleration of the proton is $4.8\times {10}^4\text{m}/{\text{s}}^2$

Part (d). The magnitude acceleration of the proton is$9.6\times {10}^4\text{m}/{\text{s}}^2$

Step-by-step solution

Step 1. Introduction

The negative sign indicates that the direction of the induced field always opposes the cause of it.

$\epsilon =-\left|\frac{d{\Phi }_{\text{m}}}{dt}\right|$

Here, ${\Phi }_{\text{m}}$is the magnetic flux linked within the loop.

If B is the magnetic field and is the A cross section area of the loop then magnetic flux${\Phi }_{\text{m}}=BA$, so the above equations becomes as follows:

$\epsilon =A\left|\frac{dB}{dt}\right|$

Comparing the two above equations:

$\oint \overrightarrow{E}\cdot \overrightarrow{d}s=A\left|\frac{dB}{dt}\right|$

For a closed curve, the integral value of a solenoid of radius becomes as follows:

$E\left(2\pi r\right)=\pi r^2\left|\frac{dB}{dt}\right|$

The equation for the induced electric field inside a loop or solenoid is given as follows

$E=\frac{r}{2}\left|\frac{dB}{dt}\right|$

Step 2. Explanation

When the magnetic field decreases, the induced current in the loop is produced in the clockwise direction and the force is in terms of the electric field is given as follows:

$F=eE$

Here, is the charge of the proton and is the electric field.

Use Newton's laws for force in terms of mass and acceleration of the proton.

$ma=eE$

The acceleration of the proton is given as follows:

$a=\frac{eE}{m}$

Substitute $\frac{r}{2}\left|\frac{dB}{dt}\right|$for E in the above equation .

$a=\left(\frac{e}{m}\right)\left(\frac{r}{2}\right)\left|\frac{dB}{dt}\right|$

Part (a)

The acceleration of the proton in the varying magnetic field is given as follows:

$a=\left(\frac{e}{m}\right)\left(\frac{r}{2}\right)\left|\frac{dB}{dt}\right|$

Substitute $1.6\times {10}^{-19}\text{C}$for $e,1.67\times {10}^{-27}\text{kg}$for $m,1.0\times {10}^{-2}\text{m}$for , and $0.10\text{T}/\text{s}$for $\left|\frac{dB}{dt}\right|$.

$\begin{array}{l}a=\left(\frac{1.6\times {10}^{-19}\text{C}}{1.67\times {10}^{-27}\text{kg}}\right)\left(\frac{0.01\text{m}}{2}\right)|0.01\text{T}/\text{s}|\\ =4.8\times {10}^4\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude acceleration of the proton is $4.8\times {10}^4\text{m}/{\text{s}}^2$. The decreasing magnetic field creates a clockwise current in that region. At point , the electric field points upward and hence the force acting on the proton directs upward. Therefore, the direction of acceleration of the proton directs upward.

Part (b)

The acceleration of the proton in the varying magnetic field is given as follows:

$\begin{array}{l}a=\left(\frac{e}{m}\right)\left(\frac{r}{2}\right)\left|\frac{dB}{dt}\right|\\ a=\left(\frac{1.6\times {10}^{-19}\text{C}}{1.67\times {10}^{-27}\text{kg}}\right)\left(\frac{0.0\text{m}}{2}\right)|0.01\text{T}/\text{s}|\\ =0\text{m}/{\text{s}}^2\end{array}$

Part (c)

(c)

The acceleration of the proton in the varying magnetic field is given as follows:

$\begin{array}{l}a=\left(\frac{e}{m}\right)\left(\frac{r}{2}\right)\left|\frac{dB}{dt}\right|\\ a=\left(\frac{1.6\times {10}^{-19}\text{C}}{1.67\times {10}^{-27}\text{kg}}\right)\left(\frac{0.01\text{m}}{2}\right)|0.01\text{T}/\text{s}|\\ =4.8\times {10}^4\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude acceleration of the proton is $4.8\times {10}^4\text{m}/{\text{s}}^2$. The decreasing magnetic field creates a clockwise current in that region. At point , the electric field points downward and hence the force acting on the proton directs downward. Therefore, the direction of acceleration of the proton directs downward.

Part (d)

d)

The acceleration of the proton in the varying magnetic field is given as follows:

$\begin{array}{l}a=\left(\frac{e}{m}\right)\left(\frac{r}{2}\right)\left|\frac{dB}{dt}\right|\\ a=\left(\frac{1.6\times {10}^{-19}\text{C}}{1.67\times {10}^{-27}\text{kg}}\right)\left(\frac{2.0\times {10}^{-2}\text{m}}{2}\right)|0.01\text{T}/\text{s}|\\ =9.6\times {10}^4\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude acceleration of the proton is$9.6\times {10}^4\text{m}/{\text{s}}^2$ . The decreasing magnetic field creates a clockwise current in that region. At point , the electric field points downward and hence the force acting on the proton directs downward. Therefore, the direction of acceleration of the proton directs downward.