91影视

According to Biot-Savart鈥檚 law, write the expression of magnetic field at point at distance $r$.

$\mathbf{B}\mathbf{=}\frac{{\mathbf{渭}}_0\mathbf{I}}{4蟺}鈭�\frac{\mathrm{dI}脳r}{{\mathbf{r}}^3}$ 鈥︹�� (1)

Here,${渭}_0$ is the permeability for free space,$I$ is the current, $r$is the distance and$dl$ is the element.

Consider the elements $d{l}_1$and $d{l}_2$at points $P(x^{\text{'}},y^{\text{'}},z^{\text{'}})$and $P^{\text{'}}(x^{\text{'}},y^{\text{'}},鈭�z^{\text{'}})$respectively.

The points $P$and $P^{\text{'}}$lie symmetrically with respect to x-y plane. Also assume a point$M(0,y,0)$ located on y-axis.

Write the expression for the magnetic field due to the elements.

$dB=\frac{{渭}_{0}I}{4蟺}\left(\frac{d{I}_1脳r_1}{r_1^3}+\frac{d{I}_2脳r_2}{r_2^3}\right)$ 鈥︹�� (2)

Here,$r_1$and$r_2$are the position vectors of point$P$and$P\text{'}$from $M$respectively.

From the above figure,

Write the expression for position vector $r_1$.

$r_1=r_M鈭�r_P$

Substitute $y\widehat{y}$for $r_M$and $x^{\text{'}}\widehat{x}+y^{\text{'}}\widehat{y}+z^{\text{'}}\widehat{z}$for $r_P$in above equation.

$\begin{array}{c}{r}_1=r_M鈭�r_P\\ =y\widehat{y}鈭�(x^{\text{'}}\widehat{x}+y^{\text{'}}\widehat{y}+z^{\text{'}}\widehat{z})\\ =鈭�x^{\text{'}}\widehat{x}+(y鈭�y^{\text{'}})\widehat{y}鈭�z^{\text{'}}\widehat{z}\end{array}$

Write the magnitude of $r_1$.

$\begin{array}{c}{r}_1=\sqrt{{(鈭�x^{\text{'}})}^2+{(y鈭�y^{\text{'}})}^2+{(鈭�z^{\text{'}})}^2}\\ =\sqrt{{x^{\text{'}}}^2+{(y鈭�y^{\text{'}})}^2+{z^{\text{'}}}^2}\end{array}$

From the above figure,

Write the expression for position vector $r_2$.

$r_2=r_M鈭�r_{P^{\text{'}}}$

Substitute$y\widehat{y}$for$r_M$and$x^{\text{'}}\widehat{x}+y^{\text{'}}\widehat{y}鈭�z^{\text{'}}\widehat{z}$ for$r_P$in above equation.

$\begin{array}{c}{r}_2=r_M鈭�r_{P^{\text{'}}}\\ =y\widehat{y}鈭�(x^{\text{'}}\widehat{x}+y^{\text{'}}\widehat{y}鈭�z^{\text{'}}\widehat{z})\\ =鈭�x^{\text{'}}\widehat{x}+(y鈭�y^{\text{'}})\widehat{y}+z^{\text{'}}\widehat{z}\end{array}$

Write the magnitude of $r_1$.

$\begin{array}{c}{r}_2=\sqrt{{(鈭�x^{\text{'}})}^2+{(y鈭�y^{\text{'}})}^2+{(鈭�z^{\text{'}})}^2}\\ =\sqrt{{x^{\text{'}}}^2+{(y鈭�y^{\text{'}})}^2+{z^{\text{'}}}^2}\end{array}$

Thus, the magnitude of $r_1$and$r_2$are equal.

$r_1=r_2=r$

Write the expression for$d{l}_1$and$d{l}_2$.

$\begin{array}{l}d{I}_1=d{x}^{\text{'}}\widehat{x}+d{y}^{\text{'}}\widehat{y}\\ d{I}_2=d{x}^{\text{'}}\widehat{x}+d{y}^{\text{'}}\widehat{y}\end{array}$

Thus, the two elements are equal.

$d{I}_1=d{I}_2=dI$

Substitute $dl$for $d{I}_1,d{I}_2$and $r$for$r_1$ and$r_2$ in equation (2)

$\begin{array}{c}dB=\frac{{渭}_{0}I}{4蟺}\left(\frac{d{I}_1脳r_1}{r_1^3}+\frac{d{I}_2脳r_2}{r_2^3}\right)\\ =\frac{{渭}_{0}I}{4蟺}\left(\frac{dI脳(r_1+r_2)}{r^2}\right)\end{array}$ 鈥︹�� (3)

As$d{l}_1$and$(r_1+r_2)$are in the same x-y plane, $dB鈥�d{I}_1脳(r_1+r_2)$is along with z axis which is perpendicular to x-y plane.

Substitute $(d{x}^{\text{'}}\widehat{x}+d{y}^{\text{'}}\widehat{y})$for $dl$, $鈭�x^{\text{'}}\widehat{x}+(y鈭�y^{\text{'}})\widehat{y}鈭�z^{\text{'}}\widehat{z}$for $r_1$, $鈭�x^{\text{'}}\widehat{x}+(y鈭�y^{\text{'}})\widehat{y}+z^{\text{'}}\widehat{z}$for $r_2$,

$\sqrt{{x^{\text{'}}}^2+{(y鈭�y^{\text{'}})}^2+{z^{\text{'}}}^2}$ for $r$in equation (3).

$\begin{array}{c}dB=\frac{{渭}_{0}I}{4蟺}\left(\frac{dI脳(r_1+r_2)}{r^2}\right)\\ =\frac{{渭}_{0}I}{4蟺}\left(\frac{(d{x}^{\text{'}}\widehat{x}+d{y}^{\text{'}}\widehat{y})脳(鈭�x^{\text{'}}\widehat{x}+(y鈭�y^{\text{'}})\widehat{y}鈭�z^{\text{'}}\widehat{z})+(鈭�x^{\text{'}}\widehat{x}+(y鈭�y^{\text{'}})\widehat{y}+z^{\text{'}}\widehat{z})}{{\left(\sqrt{{x^{\text{'}}}^2+{(y鈭�y^{\text{'}})}^2+{z^{\text{'}}}^2}\right)}^3}\right)\\ =\frac{{渭}_{0}I}{4蟺}\left(\frac{(d{x}^{\text{'}}\widehat{x}+d{y}^{\text{'}}\widehat{y})脳(鈭�2x^{\text{'}}\widehat{x}+2(y鈭�y^{\text{'}}))\widehat{y}}{\left(\sqrt{{x^{\text{'}}}^2+{(y鈭�y^{\text{'}})}^2+{z^{\text{'}}}^2}\right)}\right)\\ =\frac{{渭}_{0}I}{4蟺}\left(\frac{2(y鈭�y^{\text{'}})d{x}^{\text{'}}+2x^{\text{'}}d{y}^{\text{'}}}{{({\left(x^{\text{'}}\right)}^2+{(y鈭�y^{\text{'}})}^2+{\left(z^{\text{'}}\right)}^2)}^{\frac{3}{2}}}\right)\widehat{z}\end{array}$

From above it is clear that, the filed is running parallel to the axis of solenoid that is along z axis.

Therefore, the magnetic field runs parallel to the axis of solenoid regardless of the shape, as along the shape is constant along the length of the solenoid.