Question

A circular region in an xy plane is penetrated by a uniform magnetic field in the positive direction of the z axis. The field’s magnitude B (in Tesla) increases with time t (in seconds) according to B = at, where a is a constant. The magnitude E of the electric field set up by that increase in the magnetic field is given by Figure versus radial distance r; the vertical axis scale is set by${\mathit{E}}_{\mathbf{s}}\mathbf{=}\mathbf{300}\mathbf{}\mathbf{\mu N}\mathbf{/}\mathbf{C}$, and the horizontal axis scale is set by${\mathbf{r}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$. Find a.

Short answer

The value of constant a is , $a=3\times {10}^{-2}\mathrm{T}/\mathrm{s}$.

Step-by-step solution

Step 1: Given

$$\begin{gathered} \mathrm{B}=\mathrm{at},\mathrm{t}\mathrm{is}\mathrm{in}\mathrm{seconds}\mathrm{and}B\mathrm{is}\mathrm{in}\mathrm{Tesla} \\ {E}_s=300\mathrm{\mu N}/\mathrm{C} \\ {\mathrm{r}}_{\mathrm{s}}=4.00\mathrm{cm} \end{gathered}$$

Determining the concept

Use Faradays law of electromagnetic induction which relates line integral of electric field with rate of change of magnetic flux. As the rate of change of magnetic field is given. solve for constant.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

$$\begin{gathered} {\stackrel{,}{o}}_{ind}=\frac{d{\varphi }_B}{dt}=\oint \overrightarrow{E}.d\overrightarrow{s} \\ d{\varphi }_B=\overrightarrow{B}.d\overrightarrow{A} \end{gathered}$$

Where,${\mathrm{\Phi }}_{\mathrm{B}}$ is magnetic flux, B is magnetic field, A is area,𝜀 is emf, E is electric field.

(a) Determining the constant  

Here,

$-\frac{d{\varphi }_B}{dt}=\oint \overrightarrow{E}.d\overrightarrow{s}$

Considering the magnitude of above equation,

$$\begin{gathered} \frac{d{\varphi }_B}{dt}=\oint \overrightarrow{E}.d\overrightarrow{s}=2\tau \tau rE \\ \frac{d}{dt}\left(B\tau \tau r^2\right)=E.2\tau \tau r \\ \tau \tau r^2.\frac{dB}{dt}=E.2\tau \tau r \\ \frac{dB}{dt}=\frac{d}{dt}\left(at\right)=a \\ \tau \tau r^2.a=E.2\tau \tau r \\ \frac{a}{2}=\frac{E}{r} \end{gathered}$$

From the plot given in the problem , the slope is,

$$\begin{gathered} \frac{E_s}{r}=\frac{300\times {10}^{-6}}{2\times {10}^{-2}}=1.5\times {10}^{-2} \\ \frac{a}{2}=\frac{E}{r}=1.5\times {10}^{-2} \\ a=3\times {10}^{-2}\mathrm{T}/\mathrm{s} \end{gathered}$$

Unit can be checked from the equation, B = at

Hence, the value of constant a is , $a=3\times {10}^{-2}\mathrm{T}/\mathrm{s}$.

Therefore, the constant a is determined by using Faraday’s law and graph of electric field vs distance.