91影视

A vector field that describes the influence of magnetic nature on electric charges in motion, conductors carrying current and other materials that are characterized as magnetic is known as the magnetic field.

The magnitude of the magnetic field at the center of the segment of radius a is,

$B=\frac{{渭}_{0}l}{2蟺a}$

Here, B is the magnetic field, ${\mathrm{渭}}_0$is the permeability of free space, I is the current, and a is the radius.

The arc angle is 120潞, so each segment has equal contribution to the magnetic field which is,

$\frac{{120}^{鈭�}}{{360}^{鈭�}}=\frac{1}{3}$

So, the magnetic field contribution of one segment is,

$\begin{array}{r}B=\frac{1}{3}\frac{{渭}_{0}I}{2蟺a}\\ =\frac{{渭}_{0}I}{6蟺a}\end{array}$

Therefore, the resultant field is,

localid="1668340886545" $$\begin{gathered} \mathrm{B}={\mathrm{B}}_{20}鈭�{\mathrm{B}}_{30} \\ =\frac{{\mathrm{渭}}_0\mathrm{I}}{6{\mathrm{蟺补}}_{20}}鈭�\frac{{\mathrm{渭}}_0\mathrm{l}}{6{\mathrm{蟺补}}_{30}} \\ \mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{I}}{6\mathrm{蟺}}\left(\frac{1}{{\mathrm{a}}_{20}}鈭�\frac{1}{{\mathrm{a}}_{30}}\right) \end{gathered}$$

鈥�.. (1)

Consider the given data s below.

The current I = 12.0 A

Radius of the segment BC,${\mathrm{a}}_{30}=30\mathrm{cm}=0.300\mathrm{m}$

Radius of the segment DA,${\mathrm{a}}_{20}=20\mathrm{cm}=0.200\mathrm{m}$

Permeability of free space,${\mathrm{渭}}_0=4\mathrm{蟺}脳{10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}$

Substitute these values into equation (1).

localid="1668340988433" $$\begin{gathered} B=\frac{\left(4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}\right)\left(12.0\mathrm{A}\right)}{6蟺}\left(\frac{1}{0.200\mathrm{m}}鈭�\frac{1}{0.300\mathrm{m}}\right) \\ =4.19脳{10}^{鈭�6}\mathrm{T} \\ =4.19\mathrm{渭罢} \end{gathered}$$

Hence, the magnitude ${\mathrm{B}}_{20}$is greater than ${\mathrm{B}}_{30}$and so, the direction of the 4.19 uT magnetic field is out of the page at point P.