Mix Examples-Moving Charges and Magnetism Questions in English

(A, C, D) The net force on a charged particle moving with velocity $\vec{v}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For the velocity to remain constant,the net force must be zero $(\vec{F} = 0)$.
Case $(i)$: If $\vec{E} = 0$ and $\vec{B} = 0$,then $\vec{F} = 0$. The particle moves with constant velocity.
Case (ii): If $\vec{E} \neq 0$ and $\vec{B} \neq 0$,the forces can cancel each other if $\vec{E} = -(\vec{v} \times \vec{B})$. This is possible.
Case (iii): If $\vec{E} = 0$ and $\vec{B} \neq 0$,the force is $\vec{F} = q(\vec{v} \times \vec{B})$. If $\vec{v}$ is parallel or anti-parallel to $\vec{B}$,then $\vec{v} \times \vec{B} = 0$,so $\vec{F} = 0$. The particle moves with constant velocity.
Case (iv): If $\vec{E} \neq 0$ and $\vec{B} = 0$,the force is $\vec{F} = q\vec{E}$. For $\vec{F} = 0$,$\vec{E}$ must be zero,which contradicts the assumption. Thus,this case is not possible.
Therefore,cases $(A)$,$(C)$,and $(D)$ are possible.

(D) The centripetal force required for circular motion is provided by the electrostatic force of attraction:
$F_{\text{centripetal}} = m r \omega^{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
From this,we find the angular velocity $\omega$:
$\omega^{2} = \frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} m r^{3}} \implies \omega \propto \frac{1}{r^{3/2}}$
The moving charge $q_{1}$ constitutes a current $i$ given by:
$i = \frac{q_{1}}{T} = \frac{q_{1} \omega}{2 \pi}$
The magnetic field $B$ at the center of a circular current loop is:
$B = \frac{\mu_{0} i}{2 r} = \frac{\mu_{0}}{2 r} \left( \frac{q_{1} \omega}{2 \pi} \right) = \frac{\mu_{0} q_{1} \omega}{4 \pi r}$
Since $\omega \propto r^{-3/2}$,we have:
$B \propto \frac{\omega}{r} \propto \frac{r^{-3/2}}{r} = r^{-5/2}$
Thus,$B \propto \frac{1}{r^{5/2}}$.

(A) When a current-carrying rectangular coil is placed in a non-uniform magnetic field,the magnetic field strength varies at different points along the coil.
Since the magnetic field is non-uniform,the magnetic forces acting on different segments of the coil do not cancel each other out,resulting in a non-zero net force.
Additionally,because the forces are distributed such that they do not act through a single point or are not balanced,a non-zero net torque is also produced.
Therefore,both the net force and the net torque acting on the coil are non-zero.
Given the standard nature of such multiple-choice questions where one answer is expected,both $(a)$ and $(d)$ are physically correct statements.

(C) Consider a segment of length $\ell$ on both line charges with linear charge densities $\lambda_1$ and $\lambda_2$.
The electric force $F_E$ per unit length between the two line charges is given by Coulomb's law for infinite lines: $F_E = \frac{2 k \lambda_1 \lambda_2 \ell}{d} = \frac{2 \lambda_1 \lambda_2 \ell}{4 \pi \varepsilon_0 d} = \frac{\lambda_1 \lambda_2 \ell}{2 \pi \varepsilon_0 d}$.
The moving charges constitute currents $I_1 = \lambda_1 v$ and $I_2 = \lambda_2 v$.
The magnetic force $F_B$ per unit length between two parallel current-carrying wires is given by: $F_B = \frac{\mu_0 I_1 I_2 \ell}{2 \pi d} = \frac{\mu_0 (\lambda_1 v) (\lambda_2 v) \ell}{2 \pi d} = \frac{\mu_0 \lambda_1 \lambda_2 v^2 \ell}{2 \pi d}$.
For the magnetic attraction to balance the electric repulsion,we set $F_E = F_B$:
$\frac{\lambda_1 \lambda_2 \ell}{2 \pi \varepsilon_0 d} = \frac{\mu_0 \lambda_1 \lambda_2 v^2 \ell}{2 \pi d}$.
Simplifying the equation,we get: $\frac{1}{\varepsilon_0} = \mu_0 v^2$.
Since the speed of light $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,it follows that $c^2 = \frac{1}{\mu_0 \varepsilon_0}$,which implies $v^2 = c^2$ or $v = c$.

(D) The magnetic field at the center $O$ due to the circular loop of radius $r$ is $B_{loop} = \frac{\mu_{0} I}{2r}$ (directed outwards).
The magnetic field at the center $O$ due to the square loop of side $a$ is $B_{square} = 4 \times \frac{\mu_{0} I}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{4 \mu_{0} I}{\pi a} \times \frac{2}{\sqrt{2}} = \frac{4 \sqrt{2} \mu_{0} I}{\pi a}$ (directed inwards).
Given $a/r = 8/\pi$,so $r = \frac{\pi a}{8}$.
Substituting $r$ in $B_{loop}$,we get $B_{loop} = \frac{\mu_{0} I}{2(\pi a / 8)} = \frac{4 \mu_{0} I}{\pi a}$.
The net magnetic field $B_{net} = B_{square} - B_{loop} = \frac{4 \sqrt{2} \mu_{0} I}{\pi a} - \frac{4 \mu_{0} I}{\pi a} = \frac{4 \mu_{0} I}{\pi a} (\sqrt{2} - 1)$.
This can be rewritten as $\frac{2 \mu_{0} I}{\pi a} \times 2(\sqrt{2} - 1)$,which matches option $D$.