Moving Charges & Magnetism: Magnetic Field on the Axis of a Current-Carrying Coil

MAGNETIC INDUCTION ON THE AXIS OF A CURRENT CARRYING CIRCULAR COIL

• The objective is to derive an expression for the magnetic field induction ($B$) at a specific point ($P$) located on the axis of a circular coil that carries a steady current ($I$) using the Biot-Savart Law.

GEOMETRICAL PARAMETERS AND INITIAL SETUP

• Circular Loop Geometry:

  • Consider a circular loop with a radius denoted as $R$.
  • The loop carries a current denoted as $I$.
  • The center of the circular loop is denoted by the point $O$.

• Point of Observation:

  • Let $P$ be the point on the axis of the coil (specifically the $OX$ axis).
  • The point $P$ is located at a distance $x$ from the center $O$.

• Differential Current Element:

  • Consider an infinitesimal element of the loop with length $dl$, which carries the current $I$ in the direction of the coil.
  • The distance from the center of this element $dl$ to the observation point $P$ is denoted as $r$.
  • Based on the geometry of the system (a right-angled triangle formed by $R$, $x$, and $r$), the distance $r$ follows the Pythagorean theorem:
  • $r^2 = x^2 + R^2$

APPLICATION OF THE BIOT-SAVART LAW

• Fundamental Law Statement:

  • According to the Biot-Savart law, the magnetic induction ($dB$) at point $P$ due to the small current element $dl$ is given by the vector expression:
  • $dB = \frac{\mu_0}{4\pi} \frac{I |dl \times r|}{r^3}$

• Simplification of the Cross Product:

  • The angle between the current element vector $dl$ and the position vector $r$ is exactly $90^\circ$.
  • Consequently, the magnitude of the cross product is calculated as:
  • $|dl \times r| = dl \cdot r \cdot \sin(90^\circ) = dl \cdot r$

• Magnitude of the Differential Magnetic Field:

  • Substituting the simplified cross product back into the Biot-Savart equation yields:
  • $dB = \frac{\mu_0}{4\pi} \frac{I \cdot dl \cdot r}{r^3} = \frac{\mu_0}{4\pi} \frac{Idl}{r^2}$
  • By substituting the value for $r^2$ determined from the geometry ($r^2 = x^2 + R^2$), the expression becomes:
  • $dB = \frac{\mu_0}{4\pi} \frac{Idl}{(x^2 + R^2)}$

COMPONENT RESOLUTION AND TRIGONOMETRIC IDENTITIES

• Orientation of the Magnetic Field:

  • The differential magnetic field vector $dB$ is oriented perpendicular to the distance vector $r$.
  • To find the total field along the axis, the axial component ($dB_x$) must be isolated.

• Axial Component Calculation:

  • The component of the magnetic field along the axis is defined as:
  • $dB_x = dB \cos(\theta)$

• Defining the Trigonometric Ratio:

  • From the geometrical arrangement of the diagram, the cosine of the angle $\theta$ is determined by the ratio of the radius ($R$) to the hypotenuse ($r$):
  • $\cos(\theta) = \frac{R}{r}$
  • Given that $r = (x^2 + R^2)^{1/2}$, the ratio can be expressed as:
    • $\cos(\theta) = \frac{R}{(x^2 + R^2)^{1/2}}$

INTEGRATION AND FINAL EXPRESSION FOR MAGNETIC FIELD INDUCTION

• Combining the Expressions:

  • Substitute the expressions for $dB$ and $\cos(\theta)$ into the equation for the axial component ($dB_x$):
  • $dB_x = \left( \frac{\mu_0}{4\pi} \frac{Idl}{(x^2 + R^2)} \right) \left( \frac{R}{(x^2 + R^2)^{1/2}} \right)$
  • $dB_x = \frac{\mu_0}{4\pi} \frac{IdlR}{(x^2 + R^2)^{3/2}}$

• Integration Over the Entire Loop:

  • To find the total magnetic induction $B$, we integrate the axial component $dB_x$ over the entire length of the circular coil:
  • $B = B_x = \int dB_x$
  • $B = \frac{\mu_0 I R}{4\pi (x^2 + R^2)^{3/2}} \int dl$

• Circumference Substitution:

  • The closed integral of the differential element $dl$ over the entire circular path is equal to the circumference of the coil:
  • $\int dl = 2\pi R$

• Simplification of the Constant and Terms:

  • Substitute the circumference into the integrated expression:
  • $B = \frac{\mu_0 I R \cdot 2\pi R}{4\pi (x^2 + R^2)^{3/2}}$
  • After cancelling the terms in the numerator and denominator ($2\pi$ cancels with $4\pi$ to leave a factor of $2$), the final expression for a single loop is:
  • $B = \frac{\mu_0 I R^2}{2 (x^2 + R^2)^{3/2}}$

EXTENSION TO COILS WITH MULTIPLE TURNS

• IIf the current-carrying circular coil is composed of $N$ turns instead of just one, the magnetic field induction is multiplied by the total number of turns.
• Final Equation for N Turns:
  • $B = \frac{\mu_0 N I R^2}{2 (x^2 + R^2)^{3/2}}$" , "title": "Derivation of Magnetic Field Induction on the Axis of a Current Carrying Circular Coil"}