91Ó°ÊÓ

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

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

Formulae:

$$\begin{gathered} L=\frac{N{âˆ\dots }_B}{i} \\ {\mathrm{Î\mu }}_L=-L\frac{di}{dt} \\ {âˆ\dots }_B=BA \end{gathered}$$

The inductance of the coil:

The inductance of the inductor is

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

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

The flux through each turn when the current is increased:

The expression of the magnetic flux is

${âˆ\dots }_B=BA$

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.

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

The maximum emf across the coil:

According to the Lenz’s law,

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