Moving Charges And Magnetism – NEET UG PhysicsPractice Questions & PYQs

The magnetic field at the center of a circular current loop is determined by the formula $B = \frac{\mu_0 i}{2R}$, which scales to half that value for a semicircular arc. The semicircle residing in the x-y plane generates a field component directed along the negative z-axis, while the semicircle in the x-z plane generates an equal field component directed along the negative y-axis. Combining these two mutually perpendicular vector components of magnitude $\frac{\mu_0 i}{4R}$ via vector addition yields a resultant magnetic field magnitude of $\sqrt{\left(\frac{\mu_0 i}{4R}\right)^2 + \left(\frac{\mu_0 i}{4R}\right)^2}$, which simplifies to $\frac{\mu_0 i}{2\sqrt{2}R}$.

A sensitive galvanometer is converted into an ammeter by connecting a very small resistance in parallel, creating a shunt path that diverts excess current away from the delicate coil. Applying the principle that the potential difference across the galvanometer matches the potential difference across the parallel shunt, the relation is written as $I_g G = (I - I_g)S$. Substituting the given values into this equation gives $1.0 \times 60 = (5.0 - 1.0)S$. Solving this expression results in a shunt resistance value of $15$ ohms.

A charged particle moving in a uniform magnetic field experiences a centripetal force provided by the magnetic Lorentz force, expressed as $qvB = \frac{mv^2}{R}$. Rearranging this expression to solve for the angular frequency shows that the velocity and radius terms cancel out, leaving the frequency dependent solely on the specific charge and the magnetic field strength. Since the time period is inversely proportional to this angular frequency, the formula $T = \frac{2\pi m}{qB}$ dictates that the time period depends on neither the speed nor the radius of乨 the circular path.

The magnetic force on a moving charge is determined by the Lorentz force equation $\vec{F} = q(\vec{v} \times \vec{B})$. Substituting the given values, the cross product of velocity and the magnetic field vector becomes $(2\hat{i} + 3\hat{j}) \times 10^6 \times 2\hat{j}$, which simplifies to $4 \times 10^6 \hat{k}$ because the cross product of $\hat{j}$ with itself vanishes and $\hat{i} \times \hat{j} = \hat{k}$. Multiplying this vector by the charge of $-2\,\mu\text{C}$ yields $-8\hat{k}$, meaning a force of 8 N acts in the -z direction.

A closed current-carrying loop placed in a uniform magnetic field experiences a total net magnetic force of zero, meaning the vector sum of the forces acting on all its individual segments must always equal zero. When applying this equilibrium condition, the forces on segments PS, SR, RQ, and the unknown segment QP combine such that the perpendicular components of the force on segment QP balance the remaining forces in the system. Specifically, its horizontal component must balance the force $F_2$ from segment SR, while its vertical component must balance the net difference between $F_3$ and $F_1$. Combining these orthogonal components using the Pythagorean theorem yields the magnitude of the force on segment QP as $\sqrt{\left(F_3-F_1\right)^2+F_2^2}$.

The magnetic force acting on a moving charged particle is always perpendicular to its velocity vector, meaning the work done by the magnetic field on the particle is strictly zero, expressed mathematically as $W = \Delta K = 0$. According to the work-energy theorem, zero net work implies that the kinetic energy of the particle remains completely constant over time. Therefore, regardless of how long the particle spends inside the transverse uniform magnetic field, its kinetic energy will retain its initial value of T.

The current in a galvanometer circuit is determined by Ohm's law as $I = \frac{V}{R_G + R_1}$, where the initial total resistance is $50\,\Omega + 2950\,\Omega = 3000\,\Omega$. Using the battery voltage of 3 V, the initial current calculates to $1\,\text{mA}$ corresponding to the full-scale deflection of 30 divisions. Scaling the deflection down to 20 divisions requires the current to drop proportionally to $\frac{2}{3}\,\text{mA}$, which leads to a required total resistance of $4500\,\Omega$ and leaves a final series resistance of $4450\,\Omega$ after subtracting the internal galvanometer resistance.

An ammeter is converted to measure higher currents by placing a shunt resistor in parallel, governed by the principle that the potential drop across the galvanometer equals the potential drop across the shunt, expressed as $I_g R_G = (I - I_g)S$. Substituting the given values into this relation gives $100\text{ A} \times 13\,\Omega = (750\text{ A} - 100\text{ A})S$. Simplifying the terms results in $1300 = 650S$, which directly yields a shunt resistance value of $2\,\Omega$.

The magnetic force experienced by a moving charged particle is determined by the vector cross product of its velocity and the magnetic field, expressed as $\vec{F} = q(\vec{v} \times \vec{B})$. This results in a magnitude of $F = qvB \sin\theta$, where $\theta$ represents the angle between the two vectors. Because the sine function evaluates to zero at 0 degrees and 180 degrees, the force disappears entirely when the particle moves parallel or anti-parallel to the field. Consequently, achieving a non-zero magnetic force requires the angle to have any value other than 0 and 180 degrees.

The magnetic field at the center of a circular coil is given by $B = \frac{\mu_{0}ni}{2r}$, while the current is related to the applied potential difference through Ohm's law as $i = \frac{V}{R}$. Since the total resistance of the wire is directly proportional to its radius, substituting $R$ yields an expression where the magnetic field depends on the potential difference and the square of the radius. Equating the magnetic fields of the two coils, and accounting for the first coil having twice the radius of the second, reveals that the potential difference across the first coil must be 4 times greater than that of the second coil to maintain an identical center field.

The magnetic force acting on a moving charged particle is determined by the Lorentz force equation $\vec{F} = Q(\vec{v} \times \vec{B})$. Since the current-carrying wire produces a magnetic field directed perpendicularly inward into the page at point $P$ and the positive charge moves along the $Ox$ axis, we take the cross product of velocity and this inward magnetic field. Applying this vector product reveals that the resulting magnetic force points straight upward along the positive $Oy$ axis.

A moving electron forms a circular current loop that produces a magnetic field at its center according to $B = \frac{\mu_{0}i}{2r}$. Expressing the current as the charge per unit time given by $i = \frac{qv}{2\pi r}$, the magnetic field becomes proportional to the particle speed divided by the square of the radius. Rearranging this relationship demonstrates that the orbital radius of the electron is directly proportional to the ratio of speed to the magnetic field.

The total current required for full-scale deflection of the galvanometer is found by multiplying the number of divisions by the current per division, giving an initial current of 0.01 amperes. Converting this galvanometer into a voltmeter requires placing a high resistance in series with it to limit the current flow. Using the standard conversion formula $R = \frac{V}{I_g} - G$, substituting the voltmeter range and galvanometer resistance yields a required series resistance of 2450 ohms.

A voltmeter is designed to measure potential difference across circuit components and must be connected in parallel without drawing significant current from the main circuit. To achieve this high overall resistance, a large resistance must be connected in series with the internal galvanometer coil. This arrangement ensures that a negligible fraction of the current flows through the meter, preserving the accuracy of the potential measurement.

When a charged particle moves perpendicular to a magnetic field, the magnetic force acts perpendicular to the instantaneous velocity, resulting in zero work done ($W = 0$). Because the work-energy theorem states that net work equals the change in kinetic energy ($W = \Delta K$), the particle's speed remains entirely unchanged while it traces a uniform circular path.

The magnetic field inside a long solenoid is determined by the formula $B = \mu_{0} n i$ where $n$ represents the number of turns per unit length and $i$ represents the current. When the current is doubled and the turn density is halved, the net multiplicative effect on the magnetic field is $(2) × (\frac{1}{2})$, which equals one. Therefore, the magnitude of the magnetic field remains unchanged from its initial value.

A charged particle moving through a region containing both electric and magnetic fields experiences a combined influence known as the Lorentz force. The electric field exerts a force proportional to the electric field vector, while the magnetic field exerts a force proportional to the cross product of the velocity vector and the magnetic field vector. Combining these two independent interactions yields the total force equation $\vec{F} = q \vec{E} + q(\vec{v} \times \vec{B})$ where $q$ denotes the charge of the particle.

The magnetic field at the center of a circular coil carrying current is given by $B = \frac{\mu_{0} N I}{2 r}$ where $N$ is the number of turns and $r$ is the radius of the coil. When a fixed length of wire is reshaped from a single turn into a two-turn coil, the new radius becomes half of the original radius because the total wire length remains constant. Substituting the doubled turn number and halved radius into the magnetic field formula results in a factor of four increase compared to the initial field.

The frequency of a charged particle moving in a perpendicular magnetic field is determined by equating the magnetic force to the centripetal force, which yields the formula $f = \frac{q B}{2 \pi m}$. Substituting the charge of an electron $e$, the expression simplifies to $\frac{e B}{2 \pi m}$. Notice that the velocity and kinetic energy terms cancel out during the derivation, proving that the frequency depends solely on the particle's mass, charge, and the strength of the magnetic field.

The magnetic dipole moment of a current-carrying coil measures its maximum magnetic strength and is proportional to the current, the area enclosed by each loop, and the total number of turns. Multiplying these parameters together provides the fundamental relationship for magnetic moment expressed as $M = n i A$ where $n$ represents the number of turns, $i$ the current, and $A$ the cross-sectional area of the coil.