Moving Charges And Magnetism – NEET UG PhysicsPractice Questions & PYQs

When electric current flows through a conducting coil, it establishes a magnetic field due to the moving charge carriers while simultaneously developing an electric field between adjacent turns due to potential differences. These orthogonal electric and magnetic fields coexist continuously around the conductors of the active inductor. Consequently, the total electromagnetic energy stored within the energized coil is distributed across both field components.

Applying the right-hand rule reveals how circular magnetic fields wrap around current-carrying conductors. Along the symmetrical axis $AB$, the individual magnetic field vectors act in opposite directions with equal strength. Consequently, their vector sum becomes zero, expressed as $\vec{B}_1 + \vec{B}_2 = 0$, because the opposing components completely neutralize one another.

The magnetic field at the centre of a circular current-carrying coil is proportional to the number of turns and inversely proportional to its radius, expressed by the formula $B = \frac{\mu_0 N I}{2r}$. When a wire of length $l$ is formed into a single-turn coil, its radius is $l / (2\pi)$, leading to a magnetic field expression $B_1 = \frac{\mu_0 I}{l}$. Reforming the exact same wire length into a two-turn coil reduces the radius by half while doubling the turn count, which scales the magnetic field to $B_2 = \frac{4\mu_0 I}{l}$. Comparing these two field strengths reveals a ratio of 1 to 4.

The magnetic force per unit length between two parallel current-carrying conductors is calculated using Ampere's law, given by the formula $F = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r}$. Substituting the permeability constant factor, the unit currents, and the unit separation distance into the equation gives $F = 10^{-7} \times \frac{2 \times 1 \times 1}{1}$. This calculation directly evaluates to 2 times 10 to the power minus 7 newtons per metre.

Current division in parallel electrical paths ensures that current splits inversely to the resistance of each branch, while the net magnetic field at the center of a circular loop is found by combining the opposing magnetic contributions of its constituent arcs. Because the resistance of segment $ABC$ is half that of segment $ADC$, the current $I_1$ through $ABC$ must be twice the current $I_2$ through $ADC$. Using the total current condition $I_1 + I_2 = I_0$, these currents evaluate to $\frac{2 I_0}{3}$ and $\frac{I_0}{3}$ respectively. Each semicircular arc produces a magnetic field at the center given by $\frac{\mu_0 I}{4r}$, but since the two paths route current in opposite rotational directions, their magnetic fields oppose each other. Subtracting the smaller field from the larger one yields a net magnetic field of $\frac{\mu_0 I_0}{12 r}$ at the centre $O$.

Expressing the kinetic energy of a charged particle moving in a magnetic field through its orbital radius yields the relation $E = \frac{B^{2} q^{2} r^{2}}{2m}$. For a given magnetic field and orbital radius, the kinetic energy is inversely proportional to the mass of the particle because both particles share identical charges. A deuteron has twice the mass of a proton, meaning it requires half the kinetic energy to maintain the exact same orbital radius in the identical magnetic field, so the proton must possess twice the 50 keV energy of the deuteron to trace a 0.5 meter radius, resulting in 100 keV.

The magnetic field contribution generated by an infinitesimal current-carrying segment is determined using the Biot-Savart law, which states that the field is directly proportional to the current and the cross product of the length element with the position vector. Scaling inversely with the cube of the distance when expressed using the position vector ensures the correct inverse-square falloff of the magnetic field magnitude with distance. Combining these geometric and proportionality factors yields the exact expression $d \overset{\rightarrow}{B}= \frac{\mu_{0}}{4 \pi} i \left( \frac{\overset{\rightarrow}{d l} \times \overset{\rightarrow}{r}}{r^{3}} \right)$.

Any closed loop carrying an electric current behaves as a magnetic dipole possessing a characteristic magnetic moment vector. When placed inside a uniform magnetic field, this magnetic moment experiences a rotational influence because the field exerts forces on opposite sides of the loop in opposing directions. Consequently, a magnetic torque defined by the equation $\vec{\tau} = \vec{m} \times \vec{B}$ acts on the coil, tending to rotate it toward alignment with the field lines.

The torque acting on a current-carrying loop in a uniform magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}$, where $\vec{M}$ represents the magnetic dipole moment. In its stable equilibrium position, the loop experiences zero net torque because the magnetic moment vector aligns directly with the external magnetic field. Since the magnetic moment vector is strictly perpendicular to the surface of the coil, achieving this alignment forces the plane of the coil to orient perpendicular to the magnetic field lines.

The radius of a charged particle's circular trajectory in a uniform magnetic field is derived by equating the magnetic force to the centripetal force, yielding the relation $r = \frac{mv}{qB}$. Since the mass, charge, and magnetic field remain constant, the radius of the orbit is directly proportional to the speed of the particle. When the speed of the electrons is doubled, the radius of the circular path increases by the exact same factor. Multiplying the initial radius of 2 cm by two yields a new radius of 4 cm.

The magnetic force between two parallel current-carrying conductors is calculated using the formula $F = \frac{\mu_0 I_1 I_2 l}{2 \pi x}$, where currents flowing in the same direction naturally produce an attractive force. Substituting the permeability constant, the 10 amp currents in both wires, the 1 metre length, and the 0.1 metre separation distance into the expression gives $F = \frac{4\pi \times 10^{-7} \times 10 \times 10 \times 1}{2\pi \times 0.1}$. Simplifying this calculation yields an attractive force magnitude of 2 times 10 to the power minus 4 newtons per metre length.

The radius of a charged particle moving perpendicularly in a magnetic field is determined by balancing the magnetic force with the centripetal force, yielding the equation $r = \frac{m v}{B e}$ where velocity is found using the kinetic energy relation $K = \frac{1}{2} m v^{2}$. Substituting the given kinetic energy and the electron mass yields the particle velocity. Using this calculated velocity alongside the given magnetic induction and the elementary charge in the radius equation results in a distance of 11 cm.

Two parallel conductors carrying currents in the same direction experience an attractive magnetic force pulling them toward each other. Since wire $B$ is positioned above wire $A$, the magnetic force on wire $A$ acts in the upward vertical direction. The magnitude of this attractive magnetic force per unit length is given by the formula $\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi h}$. Substituting the given current values yields $\frac{F}{\ell} = \frac{\mu_0 I (2I)}{2\pi h}$, which simplifies to $\frac{\mu_0 I^2}{\pi h}$. For wire $A$ to remain on the floor without lifting, this upward magnetic pull must not exceed the downward gravitational weight per unit length, represented by $\lambda g$. Equating the two expressions gives $\frac{\mu_0 I^2}{\pi h} = \lambda g$, and rearranging this boundary condition isolates the minimum required height as $\frac{\mu_0 I^2}{\pi \lambda g}$.

Converting a sensitive galvanometer into an ammeter requires reducing its overall resistance and bypassing the majority of the working current safely around the delicate internal mechanism. This is achieved by connecting a very small resistance, known as a shunt, in a parallel arrangement with the galvanometer device. As a result, almost all of the excessive current flows through this low-resistance path, protecting the sensitive meter coil from thermal damage while allowing the measurement of large currents.

The magnetic force experienced by a current-carrying conductor in a uniform magnetic field is given by the relation $F = I l B \sin\theta$, where $\theta$ represents the angle between the wire's length vector and the magnetic field. For a straight wire positioned perpendicularly to the field lines, the angle satisfies $\theta = 90^\circ$, making the sine factor equal to one. Substituting the given values of current, length, and magnetic field into the formula gives $F = 1.2 \times 0.5 \times 2$, which evaluates directly to a magnitude of $1.2$ newtons.

The magnetic field at the centre of a current-carrying circular coil is calculated using the formula $B = \frac{\mu_0 N I}{2 r}$. Substituting the permeability of free space along with 1000 turns, 0.1 A current, and 0.1 m radius results in $B = \frac{4\pi \times 10^{-7} \times 1000 \times 0.1}{2 \times 0.1}$. Simplifying this calculation gives a magnetic field value of $6.28 \times 10^{-4}$ T.

The magnetic field produced by a long straight conductor at a distance is given by Ampère's law, which relates the field strength solely to the enclosed current and the radial distance from the wire. Since cross-sectional dimensions such as diameter do not appear in the external magnetic field equation, changing the wire thickness while keeping the current constant has no effect. Therefore, the magnetic field strength far away remains identical to the earlier value.

When determining the magnetic field around a current-carrying solid wire, Ampere's Circuital Law reveals that the field behaves differently inside and outside the conductor. In the region where $r < a$, the enclosed current scales with the radius, producing a linear rise represented by $B = \frac{\mu_0 I r}{2 \pi a^2}$. Beyond the surface where $r > a$, the entire current is enclosed, causing the field to decay inversely according to $B = \frac{\mu_0 I}{2 \pi r}$ and generating a plot with a straight-line ascent followed by a curved descent.

A galvanometer and its parallel shunt form a combined resistance calculated through the parallel network formula $\frac{1}{R_{eq}} = \frac{1}{G} + \frac{1}{S}$, which yields an equivalent resistance of 18/11 ohms. Applying Ohm's law with the total current of 1 amp gives the total potential difference across the parallel combination as $V = 1 \times \frac{18}{11}$ volts. Dividing this voltage by the shunt resistance of 2 ohms determines the current passing through the shunt as 9/11 amp, which is approximately 0.8 amp.

The torque acting on a current-carrying loop in a magnetic field is given by the formula $\tau = I A B \sin\theta$, where $\theta$ represents the angle between the magnetic field vector and the normal to the loop. When the plane of the loop is oriented perpendicularly to the magnetic field, the normal vector aligns parallel to the field, resulting in an angle of $\theta = 0^\circ$. Substituting the given parameters into the torque equation gives $\tau = 10 \times 0.01 \times 0.1 \times \sin(0^\circ)$, which results in a net torque of zero.