91影视

i) Diameter of the wires,$D=2.5\mathrm{mm}$

ii) Distance between the central axes, d= 20 mm

iii) Current through the wires,i =10 A

Use the formula for magnetic flux to calculate the flux in the given regions. Use the concept of symmetry to find the magnetic flux of one wire and by multiplying it by a factor of two, get the total flux in the given region. Taking the ratio of flux in the wire to the total flux of the given region, get the percentage of flux that lies inside the wire.

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:

$$\begin{gathered} 蠒=鈭�\stackrel{鈫�}{B}.d\stackrel{鈫�}{A} \\ 蔚=-\frac{d蠒}{dt} \end{gathered}$$

Where,$桅$ is magnetic flux, B is magnetic field, A is area,饾渶 is emf.

Magnetic flux per meter of wire for currents in opposite direction:

Place the wire at the origin of co-ordinate axis and wire 2 at x=L

In both the wires, the current is in opposite direction. Hence, by using the right-hand rule, the magnetic field due to the individual wires in the region between 0 to L will add up together giving net magnetic field.

In the given problem, the reflection symmetry is possible. The plane of symmetry is at $x=\frac{L}{2}$. The net field at any point in the region $0<x<\frac{L}{2}$is the same at its mirror image point L-x

Hence, by integrating over the region $0<X<\frac{L}{2}$and then multiplying by a factor of 2, get the net flux in the region $0<X<\frac{L}{2}$.

So, the magnetic flux is given by,

${蠒}_B=2{鈭�}_0^{\frac{L}{2}}{B}_{net}dA={鈭�}_0^R{B}_{net}dA+{鈭�}_R^{\frac{L}{2}}{B}_{net}dA$

As, dA=Ldx ,

${蠒}_B=2{鈭�}_0^{\frac{L}{2}}{B}_{net}dA={鈭�}_0^R{B}_{net}\left(Ldx\right)+2{鈭�}_R^{\frac{L}{2}}{B}_{net}\left(Ldx\right)$

Since, the length is constant, taking it outside the integral,

${蠒}_B/L=2{鈭�}_0^R{B}_{net}dx+2{鈭�}_R^{\frac{L}{2}}{B}_{net}dx鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(1\right)$

Here, R is the radius of the wire.

The magnetic flux inside the wire is given by equation 29-20,

$B_{in}=\left(\frac{{渭}_{0}i}{2蟿蟿R^2}\right)r$

And the magnetic field outside the wire is given by equation 29-17,

$B_{out}=\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}$

Therefore, the net magnetic field for region 0<r<R, is

$B_{net}=\left(\frac{{渭}_{0}i}{2蟿蟿R^2}\right)r+\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}$

And the net magnetic field for region $R<r<L/2$is,

$B_{net}=\left(\frac{{渭}_{0}i}{2蟿蟿r^2}\right)r+\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}$

Substituting these magnetic fields in (1 )and taking the variable r instead of x,

$\frac{{蠒}_B}{L}-2{鈭�}_0^R\left(\frac{{渭}_{0}i}{2蟿蟿R^2}r+\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}\right)dr+2{鈭�}_R^{\frac{L}{2}}\left(\frac{{渭}_{0}i}{2蟿蟿r^2}r+\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}\right)dr鈥�鈥�鈥�鈥�鈥�鈥�\left(2\right)$.

The integration of first term of equation (2) is,

$$\begin{gathered} I_1=2\frac{{渭}_{0}i}{2蟿蟿R^2}{\left[\frac{r^2}{2}\right]}_0^R+2\frac{{渭}_{0}i}{2蟿蟿}{\left[-In\left(L-r\right)\right]}_0^R \\ {I}_1=\frac{{渭}_{0}i}{2蟿蟿R^2}\left[1-2In\left(\frac{L-R}{L}\right)\right] \end{gathered}$$

$L=20mm=0.020m,R=\frac{d}{2}=\frac{0.0025m}{2}=1.25脳{10}^{-3}m,{渭}_0=4蟺脳{10}^{-7}$

Thus,

$I_1=\frac{4\mathrm{蟿蟿}脳{10}^{-7}脳10\mathrm{A}}{2\mathrm{蟿蟿}}\left[1-2\mathrm{In}\left(\frac{0.020\mathrm{m}-0.00125\mathrm{m}}{0.020\mathrm{m}}\right)\right]$

$I_1=0.23脳{10}^{-5}\mathrm{T}.\mathrm{m}$

The integration of second term of equation (2) is,

$$\begin{gathered} I_2=2{鈭�}_R^{\frac{L}{2}}\left(\frac{{渭}_{0}i}{2蟿蟿r^2}r+\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}\right)dr \\ {I}_2=2{鈭�}_R^{\frac{L}{2}}\left(\frac{{渭}_{0}i}{2蟿蟿r^2}r+\frac{{渭}_{0}i}{2蟿蟿\left(L-r\right)}\right)dr \\ {I}_2=2\frac{{渭}_{0}i}{2蟿蟿r^2}{\left[Inr\right]}_R^{\frac{L}{2}}+2\frac{{渭}_{0}i}{2蟿蟿}{\left[-In\left(L-r\right)\right]}_R^{\frac{L}{2}} \\ {I}_2=\frac{{渭}_{0}i}{蟿蟿}\left[InL/2-InR\right]-\frac{{渭}_{0}i}{蟿蟿}\left[In\left(\frac{L-L/2}{L-R}\right)\right] \\ {I}_2=\frac{{渭}_{0}i}{蟿蟿}\left[In\left(L/2R\right)-In\left(\frac{L}{2\left(L-R\right)}\right)\right] \\ {I}_2=\frac{{渭}_{0}i}{蟿蟿}\left[In\left(\frac{L-R}{R}\right)\right] \end{gathered}$$

By substituting the values,

$$\begin{gathered} {\mathrm{I}}_1=\frac{4\mathrm{蟿蟿}脳{10}^{-7}脳10\mathrm{A}}{\mathrm{蟿蟿}}\left(\mathrm{In}\frac{0.020\mathrm{m}-0.00125\mathrm{m}}{0.020\mathrm{m}}\right) \\ {\mathrm{I}}_2=1.08脳{10}^{-5}\mathrm{T}.\mathrm{m} \end{gathered}$$

Substituting in 1) ,

$$\begin{gathered} \frac{{\mathrm{蠒}}_{\mathrm{B}}}{\mathrm{L}}={\mathrm{I}}_1+{\mathrm{I}}_2 \\ \frac{{\mathrm{蠒}}_{\mathrm{B}}}{\mathrm{L}}=0.25脳{10}^{-5}\mathrm{T}.\mathrm{m}+1.08脳{10}^{-5}\mathrm{T}.\mathrm{m} \\ \frac{{\mathrm{蠒}}_{\mathrm{B}}}{\mathrm{L}}=1.3脳{10}^5\mathrm{T}.\mathrm{m} \end{gathered}$$

Hence, the magnetic flux per meter of wire that exists in the space between the two axes when the current in the wires is in opposite direction is $1.3脳{10}^5\mathrm{T}.\mathrm{m}$

Percentage of flux lies inside wire:

The first term of the relationis the flux within the wire of radius R

Hence, by taking the ratio of this flux to the total flux within the region$0<x<\frac{L}{2}$, get the percentage of the flux inside the wire.

Thus,

$\frac{0.23脳{10}^{5}T.m}{1.3脳{10}^{-5}T.m}=0.17=17\%$

Hence, percentage of flux lies inside the wires is 17%

Magnetic flux per meter of wire for currents in parallel direction:

When the currents are parallel, the magnetic field in region $0<x<\frac{L}{2}$due to one wire cancels the magnetic field due to the other wire, by using the right-hand rule.

Hence, the net magnetic flux in this region is zero.

Therefore, using the magnetic flux formula, find the flux in different regions.