Question

Two parallel coaxial circular coils of equal radius 'R' and equal number of turns 'N', carry equal currents 'I' in the same direction and are separated by a distance '2R'. Find the magnitude and direction of the net magnetic field produced at the midpoint of the line joining their centres.

Answer 1

The magnitude of the net magnetic field is $\frac{\mu_0 N I \sqrt{2}}{4 R}$. The direction is along the axis joining the centers, in the direction determined by the common direction of current in the coils.

Answer 2

The problem asks for the magnitude and direction of the net magnetic field at the midpoint of the line joining the centers of two parallel coaxial circular coils.

1. Setup and Parameters:

• Radius of each coil = $R$
• Number of turns in each coil = $N$
• Current in each coil = $I$
• Currents flow in the same direction.
• Separation between the coils = $2R$
• The point of interest is the midpoint of the line joining their centers.

2. Magnetic Field due to a Single Circular Coil on its Axis: The magnetic field ($B$) on the axis of a circular coil of radius $R'$, $N$ turns, carrying current $I$ at a distance $x$ from its center is given by: $B = \frac{\mu_0 N I R'^2}{2(R'^2 + x^2)^{3/2}}$ In our case, $R' = R$. So, $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$

3. Position of the Midpoint: Let the center of the first coil (Coil 1) be at $x=0$. Since the coils are separated by a distance $2R$, the center of the second coil (Coil 2) will be at $x=2R$. The midpoint of the line joining their centers will be at $x = \frac{0 + 2R}{2} = R$.

4. Magnetic Field due to Coil 1 at the Midpoint: For Coil 1, the distance from its center ($x=0$) to the midpoint ($x=R$) is $x_1 = R$. The magnetic field $B_1$ due to Coil 1 at the midpoint is: $B_1 = \frac{\mu_0 N I R^2}{2(R^2 + R^2)^{3/2}}$ $B_1 = \frac{\mu_0 N I R^2}{2(2R^2)^{3/2}}$ $B_1 = \frac{\mu_0 N I R^2}{2 \cdot (2^{3/2}) \cdot (R^2)^{3/2}}$ $B_1 = \frac{\mu_0 N I R^2}{2 \cdot (2\sqrt{2}) \cdot R^3}$ $B_1 = \frac{\mu_0 N I}{4\sqrt{2} R}$

5. Magnetic Field due to Coil 2 at the Midpoint: For Coil 2, the distance from its center ($x=2R$) to the midpoint ($x=R$) is $x_2 = |R - 2R| = R$. The magnetic field $B_2$ due to Coil 2 at the midpoint is: $B_2 = \frac{\mu_0 N I R^2}{2(R^2 + R^2)^{3/2}}$ $B_2 = \frac{\mu_0 N I}{4\sqrt{2} R}$ (This is the same magnitude as $B_1$ because the distance and coil parameters are identical).

6. Direction of the Magnetic Fields: Since both coils carry current in the same direction, their magnetic fields at the midpoint will also be in the same direction along the axis. For example, if the current in both coils is counter-clockwise when viewed from the right side of the coils, then the magnetic field lines would point from left to right along the axis. Thus, both $B_1$ and $B_2$ will point in the same direction.

7. Net Magnetic Field: Since $B_1$ and $B_2$ are in the same direction, the net magnetic field $B_{net}$ at the midpoint is the sum of their magnitudes: $B_{net} = B_1 + B_2$ $B_{net} = \frac{\mu_0 N I}{4\sqrt{2} R} + \frac{\mu_0 N I}{4\sqrt{2} R}$ $B_{net} = 2 \times \frac{\mu_0 N I}{4\sqrt{2} R}$ $B_{net} = \frac{\mu_0 N I}{2\sqrt{2} R}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$: $B_{net} = \frac{\mu_0 N I \sqrt{2}}{2\sqrt{2} \cdot \sqrt{2} R}$ $B_{net} = \frac{\mu_0 N I \sqrt{2}}{4 R}$

Magnitude and Direction: The magnitude of the net magnetic field is $\frac{\mu_0 N I \sqrt{2}}{4 R}$. The direction of the net magnetic field is along the axis joining the centers of the coils, in the direction determined by the common direction of current in the coils (e.g., if currents are counter-clockwise when viewed from one side, the field points away from that side).