91影视

1. Radius of loop,$r_{loop}=6cm$
2. Radius of solenoid,$r_{sol}=2cm$
3. Turn density$n=8000turns/m$
4. Vertical scale,$i_s=1A$
5. Horizontal scale,$t_s=2s$
6. Vertical scale,$E_s=100nJ$

Use the magnetic flux formula to find the flux through the area of solenoid. By substituting this value in Faraday鈥檚 law, find the magnitude of the induced emf. By using the rate of thermal energy dissipation formula, find the resistance of the coil.

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 follows:

$$\begin{gathered} B_{sol}={渭}_{0}n{i}_{sol} \\ {蠒}_B=BA \\ P=i\left|蔚\right| \end{gathered}$$

Where,role="math" localid="1661855936116" $桅$is magnetic flux, B is magnetic field, A is area, i is current, P is power, 饾渶 is emf, nis number of turns, 饾渿₀is permeability.

The magnetic field inside a solenoid is,

$B_{sol}={渭}_{0}n{i}_{sol}$

For the long solenoid, assume that the field outside the solenoid is zero and the field inside the solenoid is constant.

Then, the magnetic flux through the area of a solenoid is,

$$\begin{gathered} {蠒}_B=B_{sol}{A}_{sol} \\ {蠒}_B={渭}_{0}n{i}_{sol}脳蟿蟿r_{{}_{sol}}^2 \end{gathered}$$

Using the Faraday鈥檚 law, the magnitude of the induced emf is,

$$\begin{gathered} \left|蔚\right|=\frac{d{蠒}_B}{dt} \\ \left|蔚\right|={渭}_{0}n\frac{d{i}_{sol}}{dt}脳蟿蟿r_{sol}^2 \\ n=8000\frac{turns}{m},{渭}_0=4蟿蟿脳{10}^{-7}\frac{T.m}{A},r=0.02m \end{gathered}$$

From fig.30-53 (b) , the slope of the line gives$\frac{d{i}_{sol}}{dt}$

Therefore,

$\frac{d{i}_{sol}}{dt}=0.5A/s$

Substituting the values,

$$\begin{gathered} \left|蔚\right|=\left(4蟿蟿脳{10}^{-7}\frac{T.m}{A}\right)\left(8000turns/m\right)\left(0.5A/s\right)脳蟿蟿{\left(0.02\right)}^2 \\ \left|蔚\right|=6.3脳{10}^{-6}6.3渭V \end{gathered}$$

The rate of the dissipation of thermal energy is given as,

$P=\frac{d{E}_{th}}{dt}$

Also, from equation 26 - 28,

$P=\frac{{蔚}^2}{R}$

Thus,

$\frac{d{E}_{th}}{dt}=\frac{{蔚}^2}{R}$

Rearranging the equation for R,role="math" localid="1661856500413" $R=\frac{{蔚}^2}{\frac{d{E}_{th}}{dt}}$

From$\mathrm{Fi}g.30-53\left(c\right)$, the slope of the line gives$\frac{d{E}_{th}}{dt}$

Therefore,

$\frac{d{E}_{th}}{dt}=40nJ/s$

Thus,

$$\begin{gathered} R=\frac{{\left(6.3渭V\right)}^2}{40nJ/s} \\ R=0.992脳{10}^{-3}惟鈮�1.0m惟 \end{gathered}$$

Hence, the loop鈥檚 resistance is$1.0m惟$.

Therefore, by using the magnetic flux formula to find the flux through the area of solenoid and substituting this value in Faraday鈥檚 law, and by using the rate of thermal energy dissipation formula, the resistance of the coil can be determined.