Question

Four long, straight wires, each carrying a current of 5.0 A, are placed in a plane as shown in figure. The points of intersection form a square of side 5.0 cm.
(a) Find the magnetic field at the centre P of the square.
(b) Q1, Q2, Q3, and Q4, are points situated on the diagonals of the square and at a distance from P that is equal to the diagonal of the square. Find the magnetic fields at these points.
Figure

Answer 1

(a) The magnetic field at the centre P of the square is Zero

(b) The magnetic fields at these points

Q2 and Q4 will be zero

Q3 and Q1 will be $1.1 \times 10^{-4} T$

Explanation:

Magnetic field at P=0

(a)The magnetic field at the centre P of the square:

Consider object Q1.

Let

Horizontal wires at the bottom as well as top respectively be noted as $W_1$ and $W_2.$

Let the right and left to point P vertical wires be noted as $W_3$ and $W_4$ respectively.

Current magnitude , I = 5 A

(b) The magnetic fields at these points:

Step 1:

Consider Object P.

For size, the magnetic fields attributable to wires  $W_1$ and $W_2.$, are the same, but in the opposite direction.  

Magnetic fields have the same magnitude due to wires $W_3$ and $W_4$ but they are opposite in direction. Net magnetic field therefore is zero.

Because of these four wires net

Because of $W_1$, point $Q_1$ is separated from wire (d) by 7.5 cm.

Step 2:  

Therefore, the magnetic field is given by current in that wire  

           =$4 \times 10^{-5} T$    (In the upward direction)

Because of wire $W_2$, point $Q_1$ separation from wire (d) is 2.5 cm.  

The magnetic field is therefore supplied by the current in the wire

$B_{W 2}=\frac{4}{3} \times 10^{-5} T$      (In the upward direction)

Because of $W_3,$ point $Q_1$ is separated from wire (d) by 7.5 cm.  

So, the magnetic field in the wire is supplied with current

 $B_{W 3}=4 \times 10^{-5} T$     (In the upward direction)

Because of wire $W_4,$ point $Q_1$ from wire (d) is 2.5 cm apart.  

Thus, the magnetic field is supplied by current in the wire

$B_{W 4}=\frac{4}{3} \times 10^{-5} T$   (In the upward direction)

Step 3:

Point Q1 Net magnetic field

         $=\frac{32}{3} \times 10^{-5}$

         $=1.06 \times 10^{-5} T$   (In the upward direction) At the point Q2,

Step 5:

Magnetic field because of $W_1$ wire:

$B w_{1}=4 \times 10^{-5} \mathrm{T}$   (In the upward direction)

Magnetic field due to the wire $W_2$:

$B_{W 2}=\frac{4}{3} \times 10^{-5} T$    (In the upward direction)

Magnetic field due to  the wire $W_3$:

$B_{W 3}=\frac{4}{3} \times 10^{-5} T$   (In the downward direction)

Magnetic field due to the wire $W_4$:

$B_{W 4}=4 \times 10^{-5} T$    (In the downward direction)

Step 6:

Net magnetic field $Q_2$,

$BQ_2=0$

Likewise, the magnetic field at point  is  (in downward direction) and the magnetic field at point  is nil.

Answer 2

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