Question

A loop antenna of area$\mathbf{2}\mathbf{.}\mathbf{00}{\mathbf{cm}}^{\mathbf{2}}$and resistance 5.21 mis perpendicular to a uniform magnetic field of magnitude 17.0 mT. The field magnitude drops to zero in 2.96 ms. How much thermal energy is produced in the loop by the change in field?

Short answer

The thermal energy is produced in the loop by the change in field is $7.50\times {10}^{-10}\mathrm{J}$.

Step-by-step solution

Given

i) Area of loop$A=2{\mathrm{cm}}^2$

ii) Resistance of loop$R=5.21\mathrm{\mu }\Omega$

iii) Initial magnetic field$B_i=17\mathrm{\mu T}$

iv) Final magnetic field$B_f=0$

v) Time$∆t=2.96\mathrm{ms}$

Determining the concept

Use the rate of thermal energy dissipation formula given by theequations,to calculate the thermal energy produced in the loop. Equate both the formulae for rate of thermal energy and solved for energy. Then, substituting the given values of the area of the loop, magnetic field, resistance, and time calculate the energy generated in the loop.

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:

$$\begin{gathered} \left|\mathrm{\epsilon }\right|=\frac{d{\mathrm{\varphi }}_{\mathrm{B}}}{dt} \\ \mathrm{\varphi }=\mathrm{BA} \\ P=\frac{{\epsilon }^2}{R} \\ P=\frac{∆E_{th}}{∆t} \end{gathered}$$

Where,$\mathrm{\Phi }$is magnetic flux, B is magnetic field, A is area, R is resistance,𝜀 is emf, P is power, E is thermal energy, t is time.

Determining the thermal energy is produced in the loop by the change in field

The rate of thermal energy generated in the loop is given by the equations 26-28,

$$\begin{gathered} P=\frac{∆E_{th}}{∆t}...........................................\left(1\right) \\ P=\frac{{\epsilon }^2}{R}................................................\left(2\right) \end{gathered}$$

Where,R is the resistance of the loop.

Equating the relations (1) and (2),

$\frac{∆E_{th}}{∆t}=\frac{{\epsilon }^2}{R}.............................................\left(3\right)$

The induced emf in the loop is given by,

$$\begin{gathered} \epsilon =-\frac{d{\mathrm{\varphi }}_{\mathrm{B}}}{dt} \\ \mathrm{But},∆{\mathrm{\varphi }}_{\mathrm{B}}=∆B{A}_{loop};\mathrm{therefore} \\ \mathrm{\epsilon }=-{\mathrm{A}}_{\mathrm{loop}}\left(\frac{∆\mathrm{B}}{∆\mathrm{t}}\right) \end{gathered}$$

Thus, equation (3) becomes,

$$\begin{gathered} \frac{∆E_{th}}{∆t}=\frac{{\left(-A_{loop}\left({\displaystyle \frac{∆B}{∆t}}\right)\right)}^2}{R} \\ \mathrm{As}\left|∆B\right|=\left|B_f-B_i\right|=\left|0-17\mathrm{\mu T}\right|=\left|-17\mathrm{\mu T}\right|=17\mu T \\ ∆t=t_f-t_i=2.96\mathrm{ms}-0=2.96\mathrm{ms} \end{gathered}$$

Therefore,

$$\begin{gathered} ∆E_{th}=\frac{1}{R}{\left(-A_{loop}\frac{∆B}{∆t}\right)}^2∆t \\ ∆E_{th}=\frac{A_{loop}^2{B}^2}{R∆t} \end{gathered}$$

Substituting the values,

$$\begin{gathered} ∆E_{th}=\frac{{\left(2\times {10}^{-4}{\mathrm{m}}^2\right)}^2\left(17\times {10}^{-6}\mathrm{T}\right)}{\left(5.21\times {10}^{-6}\Omega \right)\left(2.96\times {10}^{-3}\mathrm{s}\right)} \\ ∆{\mathrm{E}}_{\mathrm{th}}=7.50\times {10}^{-10}\mathrm{J} \end{gathered}$$

Hence, the thermal energy is produced in the loop by the change in field is$7.50\times {10}^{-10}\mathrm{J}$.

Therefore, use the rate of thermal energy dissipation formula given by the equations, to calculate the thermal energy produced in the loop.