Moving Charges And Magnetism – NEET UG PhysicsPractice Questions & PYQs

The magnetic field at the centre of a flat circular coil is given by the formula $B = \frac{\mu_0 N I}{2 r}$. Substituting the permeability value alongside 50 turns, 2 A current, and 0.5 m radius yields $B = \frac{4\pi \times 10^{-7} \times 50 \times 2}{2 \times 0.5}$. Computing this expression produces a magnetic field intensity of $1.25 \times 10^{-4}$ T.

The magnetic field at the center of a closely wound circular coil depends directly on the number of turns, the current, and inversely on its radius, expressed through the relation $B = \frac{\mu_0 N I}{2R}$. Rearranging this expression to solve for the current yields $I = \frac{2RB}{\mu_0 N}$. Substituting the given values of radius, magnetic field, permeability of free space, and number of turns gives $I = \frac{2 \times 0.05 \times 3.14 \times 10^{-3}}{4\pi \times 10^{-7} \times 100}$, which evaluates to 2.5 A. The magnetic moment of a current-carrying coil is given by the product of the number of turns, current, and cross-sectional area, written as $M = N I A$ or $M = N I \pi R^2$. Substituting the calculated current and known geometric parameters into this formula results in $M = 100 \times 2.5 \times \pi \times (0.05)^2$, which calculates out to approximately 2 A m^2.

The radius of curvature for a charged particle moving in a magnetic field is determined by balancing the magnetic force with the centripetal force, expressed by the formula $R = \frac{m v}{q B \sin \theta}$. Substituting the given parameters into this expression yields $R = \frac{3 \times 10^5}{10^8 \times 0.3 \times \sin 30^\circ}$. Evaluating the denominator with the sine of 30 degrees equal to 0.5 gives 1.5 times 10 to the power 7, resulting in 0.02 meters, or 2 cm.

Ampere's law establishes that the magnetic induction surrounding an infinite straight current-carrying wire varies inversely with the radial distance from the conductor, following the relation $B \propto \frac{1}{r}$. When the distance from the wire increases from 4 cm to 12 cm, the separation triples, causing the magnetic field strength to drop to one-third of its original value. Dividing the initial induction of $10^{-3}$ tesla by three yields the new magnetic field value of $3.33 \times 10^{-4}$ tesla.

The magnetic Lorentz force acting on a moving charge is always perpendicular to its velocity vector, which means it can change the direction of motion but never do any work on the particle. When a particle moves eastward through a vertically upward magnetic field, this perpendicular force continually pulls it sideways into a circular trajectory. Because the magnitude of the force remains constant and does no work, the speed of the particle stays entirely unchanged throughout the motion.

When a charged particle travels through mutually perpendicular electric and magnetic fields with a constant velocity, the electric force acting on the particle is perfectly balanced by the magnetic Lorentz force. Equating these two opposing forces results in the defining equation $e E = e v B$. Cancelling the elementary charge from both sides isolates the velocity as the ratio of the electric field strength to the magnetic induction, given by $v = \frac{E}{B}$. Dividing the given electric field by the magnetic field directly yields a velocity of 40 ms$^{-1}$.

Electromagnetic units define the strength of a magnetic field through the force experienced by a moving charge or a current-carrying conductor in that region. The standard international measurement for this magnetic flux density assigns the name tesla to a flux density of one Weber per square meter. Consequently, whenever scientists quantify the intensity or strength of a magnetic field, the tesla serves as the foundational unit.

The magnetic force on a moving charged particle is determined by the relation $\vec{F} = q(\vec{v} \times \vec{B})$, which dictates that the force magnitude is proportional to the sine of the angle between the velocity and magnetic field vectors. When a charge travels along the positive X-direction and encounters a magnetic field directed along the negative X-direction, the angle $\theta$ between these two vectors satisfies $\theta = 180^\circ$. Because the sine of this angle evaluates to zero, the magnetic force experienced by the charge vanishes entirely, allowing it to continue moving without any deviation or acceleration.

The magnetic field produced by a long straight current-carrying conductor is inversely proportional to the perpendicular distance from the wire, expressed mathematically as $B = \frac{\mu_{0} i}{2 \pi r}$. Because the field scales reciprocally with distance, doubling the radius from $r$ to $2 r$ reduces the magnetic field strength by a factor of two. Applying this inverse relationship to the initial value gives half of 0.4 tesla, which results in 0.2 tesla.

The magnetic moment of a current-carrying circular loop is expressed as $M = IA$, showing that it scales directly with the area of the coil. Since the current remains constant through both coils, the ratio of their magnetic moments depends entirely on the ratio of their respective areas, leading to $\frac{M_1}{M_2} = \frac{\pi r_1^2}{\pi r_2^2}$. Substituting the given radii ratio into this relationship gives the square of one-half, which simplifies to the final ratio.

The circular motion of an electron in a uniform magnetic field is sustained by the magnetic Lorentz force acting as the necessary centripetal force, equating $evB = \frac{mv^2}{r}$ to determine the orbital speed as $v = \frac{eBr}{m}$. Combining this velocity expression with the given quantization condition for the magnetic flux, where $\Phi = B\pi r^2 = \frac{nh}{e}$, allows the area term $Br^2$ to be expressed as $\frac{nh}{e\pi}$. The magnetic moment of a revolving charge is defined as the product of current and area, which expands to $\mu = IA = \left(\frac{e}{T}\right)\pi r^2$ or simplifies to $\mu = \frac{evr}{2}$. Substituting the speed relation into this magnetic moment formula yields $\mu = \frac{e^2Br^2}{2m}$. Substituting the flux quantization constraint into this result gives $\mu = \frac{nh e}{2\pi m}$, which simplifies for the lowest energy state where $n = 1$ to the final expression $\frac{eh}{2\pi m}$.

A charging parallel plate capacitor with a uniformly increasing surface charge density produces a steady displacement current inside the capacitor gap that behaves identically to the conduction current in a straight wire. According to Maxwell-Ampere's law, this time-varying electric field generates an induced magnetic field that exists both between the plates and in the surrounding space outside. The magnitude of this magnetic field varies radially, starting from zero at the center axis and reaching its maximum value right at the circular peripheries of the plates before decreasing inversely with distance outside.

When a charged particle moves through a region containing both electric and magnetic fields without experiencing any deflection, the magnetic Lorentz force must be exactly balanced by the electric force. Expressing this condition vectorially requires $-e(\vec{v} \times \vec{B}) = -e\vec{E}$, which simplifies to $\vec{E} = -\vec{v} \times \vec{B}$ or ensures that the electric field vector is mutually perpendicular to both the velocity and the magnetic field vectors. Since the magnetic field is perpendicular to the electron's velocity, the electric field must act in a direction perpendicular to the magnetic field itself. The magnitude of this required electric field is given by $E = vB$. Substituting the electron speed of $c/100$, where $c$ is $3 \times 10^8\text{ ms}^{-1}$, along with the magnetic field magnitude of $9 \times 10^{-4}\text{ T}$, gives $E = \frac{3 \times 10^8}{100} \times 9 \times 10^{-4}$, resulting in a magnitude of $27 \times 10^2\text{ V m}^{-1}$.

The magnetic field at the center of a circular coil carrying current is determined using the expression $B = \frac{\mu_0 N I}{2R}$. Substituting the given values for the number of turns, current, radius, and permeability of free space yields $B = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1}$. Evaluating this calculation results in 4.4 mT after simplifying the powers of ten.

Zero deflection in the galvanometer means that the branch current $i_g = 0$, establishing a balanced condition across the network branches. Equating the current passing through the 400 ohm resistor, which experiences a potential drop of (10 - 2) volts, to the current flowing through the resistor $R$ across a 2 volt potential difference yields the relation $\frac{10-2}{400} = \frac{2}{R}$. Rearranging this equation to solve for the unknown gives $R = \frac{2 × 400}{8}$, which evaluates to 100 ohms.

When a charged particle moves parallel to a magnetic field, the magnetic force acting on it becomes zero because the cross product of its velocity and the magnetic field is zero, which can be represented as $F_B = 0$. However, the uniform electric field exerts an electrostatic force on the electron opposite to its direction of motion due to its negative charge. As a result, this opposing force does negative work on the electron, causing its speed to decrease as it moves forward.

Equating the magnetic force acting on a current-carrying conductor to its weight provides the balance condition given by $mg = ILB$. Inserting the specified mass, acceleration due to gravity, length, and magnetic field strength into this equation sets up $0.250 \times 9.8 = i \times 2 \times 0.7$. Solving this expression for the current yields 1.75 A after isolating the variable.

The magnetic force on a current-carrying conductor is given by the vector equation $\vec{F} = I(\vec{L} \times \vec{B})$. Substituting the length vector along the positive x-axis and the given magnetic field components yields a cross product that simplifies into orthogonal vector components. Taking the magnitude of this resulting force vector gives $\sqrt{\left(\frac{3}{L}\right)^2 + \left(\frac{4}{L}\right)^2}$. Simplifying this square root expression leads directly to the final magnitude of $\frac{5}{L}$.

Determining the magnetic field at point $P$ requires applying the Biot-Savart law through the principle of superposition, treating the bent wire as three distinct segments made up of two semi-infinite straight portions and one semicircular arc. Each semi-infinite straight wire generates a magnetic field of magnitude $\frac{\mu_0 i}{4 \pi R}$ at the center, and applying the right-hand rule reveals that both of these straight sections direct their magnetic fields directly into the page.

Meanwhile, the semicircular portion of radius $R$ produces a magnetic field of magnitude $\frac{\mu_0 i}{4 R}$ at its center, with the right-hand rule indicating that this contribution points outward, away from the page. Combining these opposing directional effects by taking the outward semicircular field and subtracting the combined inward fields from the two straight segments yields a net magnetic field pointing away from the page with a magnitude of $\frac{\mu_0 i}{4 R} \left[1 - \frac{2}{\pi}\right]$.

The magnetic field at an axial distance from the center of a circular current-carrying loop is determined by the Biot-Savart law integrated over the loop, resulting in the standard formula $B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}.$ Substituting the given radius $R = 1$, distance $x = 1$, and current $I = \sqrt{2}$ into this expression yields $B = \frac{4\pi \times 10^{-7} \times \sqrt{2} \times 1^2}{2(1^2+1^2)^{3/2}}.$ Simplifying the denominator gives $2(2\sqrt{2}) = 4\sqrt{2}$, which cancels the $\sqrt{2}$ \in the numerator along with the numerical factors to leave $\pi × 10^{-7}$, evaluating to $3.14 × 10^{-7}$ teslas.