Question

The magnetic field of a cylindrical magnet that has a pole-face diameter of 3.3 cmcan be varied sinusoidally between 29.6 Tand 30.0 Tat a frequency of 15Hz. (The current in a wire wrapped around a permanent magnet is varied to give this variation in the net field.) At a radial distance of 1.6 cm, what is the amplitude of the electric field induced by the variation?

Short answer

Amplitude of electric field induce is,$E_m=0.15\frac{\mathrm{V}}{\mathrm{m}}$

Step-by-step solution

Step 1: Given

$r=1.6\times {10}^{-2}\mathrm{m}$

Diameter of cylindrical magnet $D=3.3\mathrm{cm}=3.3\times {10}^{-2}\mathrm{m}$

Frequency of sinusoidally varying magnetic field, f = 15.0 Hz

Determining the concept

The rate of change of magnetic flux is related to the line integral of the electric field by Faraday’s law, so use Faraday’s law to evaluate the line integral,$\mathbf{\oint }\overrightarrow{\mathbf{E}}\mathbf{.}\mathit{d}\overrightarrow{\mathbf{s}}$. Using this, find the magnitude of the electric field inside the solenoid at distance rfrom the axis of the solenoid.

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:

$$\begin{gathered} {\ddot{o}}_{ind}=-\frac{d{\varphi }_B}{dt}=\oint \overrightarrow{E}.d\overrightarrow{s} \\ d{\varphi }_B=\overrightarrow{B}.d\overrightarrow{A} \\ E=\frac{r}{2}\frac{dB}{dt} \\ \omega =2\tau \tau f \end{gathered}$$

Where,${\mathbf{\Phi }}_{\mathbf{B}}$is magnetic flux, B is magnetic field, A is area,𝜀 is emf, E is electric field,𝜔 is angular velocity, f is frequency, r is radius.

Determining the amplitude of electric field induced

As mentioned in the problem the magnetic field is varying sinusoidally, let’s assume that the following form of time is dependent on magnetic field,

$B=B_0\sin \left(\omega t\right)=B_0\sin \left(2\tau \tau ft\right)$

As the field is varying in between two values 30.0 T and 29.6 T

Taking difference and dividing it by 2, the amplitude of field,

$$\begin{gathered} B_0=\left(\frac{30.0-29.6}{2}\right)=\frac{04}{2}=0.2T \\ B=\left(0.2T\right)\sin \left(2\tau \tau \times 15\times t\right)=\left(0.2T\right)\sin \left(30\tau \tau t\right) \end{gathered}$$

Now, consider magnitude electric field inside solenoid at distance r from the axis of solenoid is,

$$\begin{gathered} E=\frac{r}{2}\frac{dB}{dt}=\frac{r}{2}.\frac{d}{dt}\left\{\left(02.T\right)\sin \left(30\tau \tau t\right)\right\}=0.2\times \frac{r}{2}\times \left(30\tau \tau \right)\cos \left(30\tau \tau t\right) \\ E=0.2\times \frac{1.6\times {10}^{-2}}{2}\times \left(30\tau \tau \right)\cos \left(30\tau \tau t\right) \end{gathered}$$

E will be maximum when cosine will have maximum value which is equal to one

$$\begin{gathered} E_{max}=0.2\times \frac{1.6\times {10}^{-2}}{2}\times \left(30\tau \tau \right) \\ {E}_{max}=15.072\times {10}^{-2} \\ {E}_{max}=0.15\frac{\mathrm{V}}{\mathrm{m}} \end{gathered}$$

Hence, amplitude of electric field induce is,$E_{max}=0.15\frac{\mathrm{V}}{\mathrm{m}}$

Therefore, value cosine function fluctuates between +1 and -1. The magnitude of cosine cannot exceed one, by considering this fact, the amplitude of electric field induced can be found.