Question

A coil with 150turns has a magnetic flux of $\mathbf{50}\mathbf{.}\mathbf{0}\mathit{n}\mathit{T}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$ through each turn when the current is 2.00mA . (a) What is the inductance of the coil? What are the (b) inductance and (c) flux through each turn when the current is increased to i = 4.00mA ? (d) What is the maximum emf across the coil when the current through it is given by i= (3.00mA)cos(377 t) , with t in seconds?

Short answer

1. The inductance of the coil is L = 3.75mH
2. The value of the inductance when the current is increased is L= 3.75mH
3. The flux through each turn when the current is increased as${\varnothing }_B=100\mathrm{nT}.{\mathrm{m}}^2$
4. The maximum emf across the coil is,${\epsilon }_{Lmax}=4.24\times {10}^{-3}\mathrm{V}$

Step-by-step solution

Given

1. The number of turns of the coil is N = 150
2. The magnetic flux of the coil is ${\varnothing }_{\mathrm{B}}=50.0\mathrm{nT}.{\mathrm{m}}^2=50.0\times {10}^{-9}\mathrm{T}.{\mathrm{m}}^2$
3. The current passing through the coil is$i=2.00mA=2.00\times {10}^{-3}A$
4. The increasing current in the coil is$i=4.00mA=4.00\times {10}^{-3}A$
5. The current through the coil is $i=(3.00mA)\cos \left(377t\right)$

Understanding the concept

We can use the concept of the inductance of the inductor and Lenz’s law. We can use the expression of magnetic flux.

Formulae:

$$\begin{gathered} L=\frac{N{\varnothing }_B}{i} \\ {\epsilon }_L=-L\frac{di}{dt} \\ {\varnothing }_B=BA \end{gathered}$$

(a) Calculate the inductance of the coil

The inductance of the coil:

The inductance of the inductor is

$$\begin{gathered} L=\frac{N{\varnothing }_B}{i} \\ L=\frac{150\times 50.0\times {10}^{-9}\mathrm{T}.{\mathrm{m}}^2}{2.00\times {10}^{-3}\mathrm{A}} \\ L=3.75\times {10}^{-3}\mathrm{H} \\ L=3.75\mathrm{mH} \end{gathered}$$

(b) Calculate the value of the inductance when the current is increased

The value of the inductance when the current is increased:

Due to changing the current with time, an emf is induced in the coil. According to the Lenz’s law, the self-induced emf acts to oppose the change of the current with time and inductance is constant. Hence, it remains the same.

L = 3.75mH

(c) Calculate the flux through each turn when the current is increased

The flux through each turn when the current is increased:

The expression of the magnetic flux is

${\varnothing }_B=BA$

The magnetic flux depends upon the magnetic field. The magnetic field is directly proportional to the current passing through the coil. The current is increased by double, then the magnetic flux also increases by two.

$$\begin{gathered} {\varnothing }_B=2(50.0\times {10}^{-9}\mathrm{T}.{\mathrm{m}}^2) \\ {\varnothing }_B=100\times {10}^{-9}\mathrm{T}.{\mathrm{m}}^2 \end{gathered}$$

(d) Calculate the maximum emf across the coil

The maximum emf across the coil:

According to the Lenz’s law,

$$\begin{gathered} {\epsilon }_{Lmax}=-\mathrm{L}\frac{\mathrm{di}}{{\mathrm{dt}}_{\max }} \\ {\epsilon }_{Lmax}=-(3.75\times {10}^{-3}\mathrm{H})\frac{\mathrm{d}\left[(3.00\mathrm{mA})\cos \left(377\mathrm{t}\right)\right]}{\mathrm{dt}} \\ {\epsilon }_{Lmax}=-(3.75\times {10}^{-3}\mathrm{H})\times (3.00\times {10}^{-3}\mathrm{A})\times 377\mathrm{rad}/\mathrm{s}\times {\left[\sin \left(377\mathrm{t}\right)\right]}_{\max } \\ {\epsilon }_{Lmax}=-4.24\times {10}^{-3}\mathrm{V} \end{gathered}$$