विद्युतचुंबकत्व

The magnetic field inside an ideal long solenoid is given by:

Given values:

• Number of turns $N = 800$
• Length $L = 0.3\,\text{m}$
• Current $I = 6\,\text{A}$
• Permeability $\mu_0 = 4\pi \times 10^{-7}\,\text{T}\cdot\text{m}/\text{A}$
  Calculating $B$:
  $$B = (4\pi \times 10^{-7}) \times \left(\frac{800}{0.3}\right) \times 6 = (4\pi \times 10^{-7}) \times 16000 = 6.4\pi \times 10^{-3}\,\text{T}$$
  $$B = 6.4 \times 3.1416 \times 10^{-3} \approx 20.1 \times 10^{-3}\,\text{T} \approx 20\,\text{mT}$$

• Current division in parallel branches:
  The two semicircular branches ABC (resistance $2R$) and ADC (resistance $R$) are in parallel.
  Current divides inversely to resistance:
  $$I_1 (\text{along ABC}) = 3I \times \frac{R}{2R + R} = I$$
  $$I_2 (\text{along ADC}) = 3I \times \frac{2R}{2R + R} = 2I$$
• Magnetic field at the center of a semicircular arc carrying current is $B = \frac{\mu_0 I}{4r}$:
• The current $I_1 = I$ in ABC flows clockwise $\rightarrow \vec{B}_1 = \frac{\mu_0 I}{4r}$ into the plane ($\otimes$).
• The current $I_2 = 2I$ in ADC flows anticlockwise $\rightarrow \vec{B}_2 = \frac{\mu_0 (2I)}{4r}$ out of the plane ($\odot$).
• Net magnetic field at the center:
  $$\vec{B}_{\text{net}} = \vec{B}_2 - \vec{B}_1 = \frac{2\mu_0 I}{4r} - \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r}\quad (\text{directed out of the plane})$$