Solution :
\( B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.1} = 10^{-5} \, \text{T} \) (soit 10 μT)

Solution :
\( F = I \cdot L \cdot B \cdot \sinθ = 3 \times 0.2 \times 0.5 \times 1 = 0.3 \, \text{N} \)
Direction : perpendiculaire au plan (règle de la main droite)

Solution :
\( B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.05} = 2.51 \times 10^{-5} \, \text{T} \)

Solution :
\( F = |q| \cdot v \cdot B \cdot \sin90° = 1.6 \times 10^{-19} \times 10^6 \times 0.1 = 1.6 \times 10^{-14} \, \text{N} \)
Sens : selon la règle de la main droite (négatif ⇒ sens inverse)

Solution :
\( B = \mu_0 \cdot n \cdot I = 4\pi \times 10^{-7} \times 1000 \times 0.5 = 6.28 \times 10^{-4} \, \text{T} \)

Solution :
\( \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 10 \times 10}{2\pi \times 0.2} = 10^{-4} \, \text{N/m} \)
Attractive car courants de même sens

Solution :
\( r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 10^5}{1.6 \times 10^{-19} \times 0.2} = 5.22 \times 10^{-3} \, \text{m} \) (5.22 mm)

Solution :
\( m = N \cdot I \cdot S = N \cdot I \cdot \pi R^2 \)
\( N = \frac{m}{I \pi R^2} = \frac{1.41 \times 10^{-3}}{0.5 \times \pi \times (0.03)^2} \approx 100 \, \text{spires} \)

Solution :
\( B_{\text{fil}} = B_{\text{terre}} \) ⇒ \( \frac{\mu_0 I}{2\pi r} = 20 \times 10^{-6} \)
\( I = \frac{20 \times 10^{-6} \times 2\pi \times 0.1}{4\pi \times 10^{-7}} = 10 \, \text{A} \)

Solution :
1. \( r = \frac{mv}{qB} = \frac{6.64 \times 10^{-27} \times 10^6}{2 \times 1.6 \times 10^{-19} \times 1} = 2.075 \times 10^{-2} \, \text{m} \) (2.08 cm)
2. \( f = \frac{qB}{2\pi m} = \frac{3.2 \times 10^{-19} \times 1}{2\pi \times 6.64 \times 10^{-27}} = 7.67 \times 10^6 \, \text{Hz} \)
3. \( E_c = \frac{1}{2}mv^2 = \frac{1}{2} \times 6.64 \times 10^{-27} \times (10^6)^2 = 3.32 \times 10^{-15} \, \text{J} \approx 20.75 \, \text{keV} \)