Question

State Biot-Savart law. Find the expression for the magnetic field due to a current carrying circular loop at its centre. Also, write down the formula for the magnetic moment of this current loop.

Hint:

The magnetic field at the center of a circular current loop is \( B = \frac{\mu_0 I}{2r} \), and the magnetic moment is \( \mu = I \cdot \pi r^2 \).

Answer 1

Step 1: Biot-Savart Law.
The Biot-Savart law gives the expression for the magnetic field produced by a small current element. It is given by: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2} \] where: - \( d\vec{B} \) is the infinitesimal magnetic field produced by the current element, - \( I \) is the current through the conductor, - \( d\vec{l} \) is the infinitesimal length vector of the current element, - \( \hat{r} \) is the unit vector from the current element to the point of observation, - \( r \) is the distance from the current element to the point of observation, - \( \mu_0 \) is the permeability of free space.
Step 2: Magnetic Field Due to a Current Carrying Circular Loop.
The magnetic field at the center of a circular loop of radius \( r \), carrying a current \( I \), is given by the formula: \[ B = \frac{\mu_0 I}{2r} \] This expression is derived using the Biot-Savart law by integrating over the entire loop.
Step 3: Magnetic Moment of the Current Loop.
The magnetic moment \( \mu \) of a current loop is given by: \[ \mu = I \cdot A \] where: - \( I \) is the current, - \( A \) is the area of the loop. For a circular loop, the area \( A = \pi r^2 \), so the magnetic moment becomes: \[ \mu = I \cdot \pi r^2 \]