Electronic l2

Step 1: State the Biot-Savart law.
The infinitesimal magnetic field $dB$ due to a current element $I \, dl$ at distance $r$ is
$dB = \frac{\mu_0}{4\pi} \frac{I \, dl \sin \theta}{r^2}$
where $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T \cdot m / A}$ (units: tesla-meters per ampere), $I$ in amperes (A), $dl$ in meters (m), $r$ in meters (m), $\theta$ is the angle between current direction and line to point.

Step 2: Apply to center of circular loop of radius $R$.
At the center, $r = R$ for all elements, $\theta = 90^\circ$ so $\sin \theta = 1$, and all $d\mathbf{B}$ components add along the axis (perpendicular to plane).
Magnitude:
$dB = \frac{\mu_0}{4\pi} \frac{I \, dl}{R^2}$

Step 3: Express $dl$ for loop.
Arc length $dl = R \, d\phi$, where $\phi$ from $0$ to $2\pi$.
Substitute:
$dB = \frac{\mu_0}{4\pi} \frac{I (R \, d\phi)}{R^2} = \frac{\mu_0 I \, d\phi}{4\pi R}$

Step 4: Integrate over full loop.
$B = \int dB = \frac{\mu_0 I}{4\pi R} \int_0^{2\pi} d\phi = \frac{\mu_0 I}{4\pi R} \cdot 2\pi = \frac{\mu_0 I}{2 R}$

Step 5: Verify units.
$\mu_0$ in $\mathrm{T \cdot m / A}$, $I$ in A, $R$ in m:
$B = \frac{T \cdot m / A \cdot A}{m} = \mathrm{T}$ (tesla).

Equipment: Power supply, ammeter, voltmeter, circular loop (radius $R$), Hall probe or teslameter, retort stand, connecting wires.

Investigation method:
Step 1: Set up circuit with power supply, ammeter in series with loop. Mount loop vertically, place Hall probe at center.
Step 2: Adjust current $I$ from 0.5 A to 5.0 A in steps of 0.5 A, record $B$ each time. Repeat for accuracy.
Step 3: Plot graph of $B$ vs $I$.

Write down your findings:
The magnetic field $B$ at the center is directly proportional to current $I$: $B \propto I$.
Gradient of graph = $\frac{\mu_0}{2R}$. Measure $R$, calculate $\mu_0$.
Sample data (example):
$I = 1.0 \, \mathrm{A}$, $B = 3.14 \times 10^{-5} \, \mathrm{T}$ (for $R=0.05 \, \mathrm{m}$).
Formula:
$B = \frac{(4\pi \times 10^{-7}) (1.0)}{2(0.05)} = 1.26 \times 10^{-5} \, T$ (adjusts with measured $R$).
Graph linear through origin.
Final formula:
B = \frac{\mu_0 I{2 R}}