Question

Figure shows an arrangement in which long parallel wires carrying equal currents (I) into or out of the page of paper at the corners of a square. The magnetic field at the centre of square is $\frac{n\mu_oI}{2\pi r}$ the value of n is

Answer 1

4

Answer 2

The problem asks us to find the value of 'n' in the expression for the magnetic field at the center of a square, due to four long parallel wires carrying equal currents (I) at its corners. The given expression for the magnetic field is $B_{total} = \frac{n\mu_0I}{2\pi r}$, where 'r' is the distance from each corner to the center of the square.

Let's denote the corners of the square as follows:

• Top-Left (TL): Current 'I' into the page (represented by X). Let's call this wire A.
• Top-Right (TR): Current 'I' out of the page (represented by •). Let's call this wire B.
• Bottom-Right (BR): Current 'I' into the page (represented by X). Let's call this wire C.
• Bottom-Left (BL): Current 'I' out of the page (represented by •). Let's call this wire D.

The center of the square is O. The distance from each wire (A, B, C, D) to the center O is given as 'r'.

The magnetic field (B) produced by a long straight wire carrying current 'I' at a distance 'd' from it is given by:

$B = \frac{\mu_0 I}{2\pi d}$

In this case, for each wire, the distance to the center O is 'r'. So, the magnitude of the magnetic field produced by each wire at the center O is the same:

$B_A = B_B = B_C = B_D = B_{individual} = \frac{\mu_0 I}{2\pi r}$

Now, we need to determine the direction of the magnetic field produced by each wire at the center O using the right-hand thumb rule. Point your thumb in the direction of the current, and your fingers curl in the direction of the magnetic field lines. The magnetic field vector at a point is tangential to the magnetic field line at that point.

Let's visualize the directions:

1. Wire at A (Top-Left, Current In 'X'):

   • Place your right thumb pointing into the page at A. Your fingers curl clockwise.
   • At the center O, the magnetic field $B_A$ will be perpendicular to the line AO and directed clockwise around A. This means $B_A$ points towards the bottom-right corner (C).

2. Wire at B (Top-Right, Current Out '•'):

   • Place your right thumb pointing out of the page at B. Your fingers curl counter-clockwise.
   • At the center O, the magnetic field $B_B$ will be perpendicular to the line BO and directed counter-clockwise around B. This means $B_B$ also points towards the bottom-right corner (C).

3. Wire at C (Bottom-Right, Current In 'X'):

   • Place your right thumb pointing into the page at C. Your fingers curl clockwise.
   • At the center O, the magnetic field $B_C$ will be perpendicular to the line CO and directed clockwise around C. This means $B_C$ also points towards the bottom-right corner (C).

4. Wire at D (Bottom-Left, Current Out '•'):

   • Place your right thumb pointing out of the page at D. Your fingers curl counter-clockwise.
   • At the center O, the magnetic field $B_D$ will be perpendicular to the line DO and directed counter-clockwise around D. This means $B_D$ also points towards the bottom-right corner (C).

All four magnetic field vectors ($B_A, B_B, B_C, B_D$) at the center O point in the same direction (towards the bottom-right corner C). Since they are all in the same direction and have the same magnitude, the net magnetic field at the center O is the sum of their magnitudes.

$B_{total} = B_A + B_B + B_C + B_D$ $B_{total} = B_{individual} + B_{individual} + B_{individual} + B_{individual}$ $B_{total} = 4 \times B_{individual}$ $B_{total} = 4 \times \frac{\mu_0 I}{2\pi r}$ $B_{total} = \frac{4\mu_0 I}{2\pi r}$

Comparing this with the given expression $B_{total} = \frac{n\mu_0I}{2\pi r}$, we can see that the value of $n$ is 4.