91影视

The magnetic field due to the very long wire is given by

$B=\frac{{渭}_{0}l}{2\mathrm{蟺谤}}$

Where, is the magnetic field due to wire,${渭}_0$ is the permeability of the vaccum, l is the current through the wire, and r is the distance from the wire.

The direction of the magnetic field due to the current-carrying wire:

The direction of the magnetic field due to the current-carrying conductor can be given by the right-hand thumb rule.

According to the right-hand thumb rule, if the thumb of the right-hand points along the direction of the current, then the remaining curled fingers of the same hand give the direction of the magnetic field due to the current.

Given data:

• The current flowing through the wire is =100 A.
• Side of square is = 20.0 cm

Using Pythagoras theorem as;

$\begin{array}{r}r=\sqrt{(10\mathrm{cm}{)}^2+(10\mathrm{cm}{)}^2}\\ r=10\sqrt{2}\mathrm{cm}\end{array}$

Where, r is the distance between center and wire.

a)

Calculation of magnetic field in the first case

Here, The magnetic field due to the first wire and the magnetic field due to the third wire is the same in magnitude but opposite in direction (Clockwise).

Similarly, the magnetic field due to the second wire and the magnetic field due to the fourth wire is the same in magnitude but opposite in direction (Clockwise).

Hence, the magnetic field at the center of the square is B = 0 T .

b)

Calculation of magnetic field in the second case

Here, the magnetic field due to the first wire and the magnetic field due to the third wire is the same in magnitude but opposite in direction(Clockwise).

Similarly, the magnetic field due to the second wire and the magnetic field due to the fourth wire is the same in magnitude but opposite in direction(Anticlockwise).

Hence, the magnetic field at the center of the square is B = 0 T.

c)

Calculation of magnetic field in the third case

Here, the magnetic field due to the first wire and the magnetic field due to the third wire are the same in magnitude same in direction.

So, net magnetic field is:

$\begin{array}{r}{B}^{鈥测赌�}=B_1+B_3\\ B^{鈥测赌�}=2B_0\end{array}$

Where B'' is the net magnetic field due to the first and third wire, and $B_0$is the magnetic field due to the first and third wire

Similarly, the magnetic field due to the second wire and the magnetic field due to the fourth wire is the same in magnitude but opposite in direction.

So, net magnetic field is:

$\begin{array}{r}{B}^{\mathrm{鈥�}}=B_2+B_4\\ B^{\mathrm{鈥�}}=2B_0\end{array}$

Where B' is the net magnetic field due to the first and third wire, and $B_0$is the magnetic field due to the second and fourth wire.

Here, the angle between two vectors B' and B'' is $90掳$.

So, according to vector summation of B' and B'' :

• Vertical components cancel each other.
• Horizontal components are added together.

$\begin{array}{r}\stackrel{鈫�}{B_{\text{net}}}=B鈰�\cos 鈦�{45}^{鈭�}(鈭�\widehat{i})+B鈰�\cos 鈦�{45}^{鈭�}(鈭�\widehat{i})\\ \stackrel{鈫�}{B_{\text{net}}}=鈭�2\left(2B_0\cos 鈦�{45}^{鈭�}\right)\widehat{i}\\ \stackrel{鈫�}{B_{\text{net}}}=鈭�4脳\left(\frac{{渭}_{0}l}{2蟺r}\right)脳\cos 鈦�{45}^{鈭�}\widehat{i}\end{array}$

Putting the values of constants in the above equation

$\begin{array}{r}{\stackrel{鈫�}{B}}_{\text{net}}=鈭�4脳\left(\frac{4蟺脳{10}^{鈭�7}脳100}{2蟺脳10\sqrt{2}脳{10}^{鈭�2}}\right)脳\cos 鈦�{45}^{鈭�}\widehat{i}\\ {\stackrel{鈫�}{B}}_{\text{net}}=鈭�4.0脳{10}^{鈭�4}\widehat{i}\mathrm{T}\end{array}$

Thus, the magnetic field is $4.0脳{10}^{鈭�4}\mathrm{T}$towards the left of the center.