विद्युतचुंबकत्व

• When a particle of mass $m$ and charge $q$ is accelerated from rest through potential difference $V$:
  $$q V = \frac{1}{2} m u^2 \implies \frac{q}{m} = \frac{u^2}{2V}$$
• In a perpendicular uniform magnetic field $B$, the magnetic Lorentz force provides the necessary centripetal force:
  $$q u B = \frac{m u^2}{r} \implies r = \frac{m u}{q B}$$
• Substituting $\frac{m}{q} = \frac{2V}{u^2}$:
  $$r = \frac{u}{B} \left(\frac{2V}{u^2}\right) = \frac{2V}{B u}$$
• Since $B$ is constant, the radius is directly proportional to $V/u$:
  $$r \propto \frac{V}{u}$$

The magnetic field produced by an infinitely long straight wire carrying current $I_1$ at a distance $d$ is:

The magnetic force exerted on a segment of length $L$ of the second parallel conductor carrying current $I_2$ is:

• 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:
  $$\vec{B}_{\text{net}} = \vec{B}_2 - \vec{B}_1 = \frac{2\mu_0 I}{4r} - \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r}\quad (\text{directed out of the plane})$$