Moving Charges And Magnetism – NEET UG PhysicsPractice Questions & PYQs

The radius of a charged particle moving perpendicularly in a uniform magnetic field is expressed as $R = \frac{\sqrt{2m(KE)}}{qB}$. Equating the radii for the proton and the alpha particle shows that kinetic energy is directly proportional to the square of the charge and inversely proportional to the mass, meaning $KE \propto \frac{q^2}{m}$. Given that an alpha particle has four times the mass and twice the charge of a proton, its ratio of $q^2/m$ is identical to that of the proton. Consequently, the alpha particle possesses the exact same kinetic energy of 1 MeV.

Calculating the total magnetic field at the origin requires applying the Biot-Savart law by dividing the wire into three separate geometric segments: two very long straight parallel sections and one semicircular arc. The net magnetic field is obtained through vector superposition of the magnetic fields produced by each individual section.

Each ofзел the semi-infinite straight wires running parallel to the X-axis contributes a magnetic field directed along the negative Z-axis at the center of the arc. Using the magnetic field formula for a semi-infinite wire, the contribution from each of these straight parts is expressed as $\vec{B}_1 = \vec{B}_3 = -\frac{\mu_0 I}{4 \pi R} \hat{k}$.

The semicircular portion of radius $R$ lying in the Y-Z plane generates a magnetic field directed along the negative X-axis, given by $\vec{B}_2 = -\frac{\mu_0 I}{4 R} \hat{i}$. Summing these vector components together yields the comprehensive magnetic field expression of $\vec{B} = -\frac{\mu_0}{4\pi} \frac{I}{R}(\pi\hat{i}+2\hat{k})$.

The magnetic field at the centre of a circular current loop is determined by the Biot-Savart law, expressed as $B = \frac{\mu_0 I}{2r}$. The equivalent current for a single revolving electron is the product of its charge and frequency, giving $I = ne$. Substituting this current into the magnetic field formula yields $B = \frac{\mu_0 ne}{2r}$.

An ammeter consists of a galvanometer connected in parallel with a shunt resistor, where the total resistance is given by the parallel combination $R_A = \frac{GS}{G+S}$. Since only $0.2\%$ of the main current passes through the galvanometer, the ratio of total current to galvanometer current evaluates to $n = \frac{I}{I_g} = 500$. Substituting this multiplying factor into the ammeter resistance formula gives $R_A = \frac{G}{500}$.

The magnetic field produced by a long straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$. Because the two identical wires are arranged at right angles to each other, the magnetic fields they generate at point P act along mutually perpendicular directions. Applying the principle of superposition, the net magnetic field is obtained by taking the vector sum of these orthogonal components, resulting in $B = \frac{\mu_0}{2\pi d}(I_1^2 + I_2^2)^{1/2}$.

An electron moving through a magnetic field experiences a magnetic Lorentz force whose magnitude is given by $F = qvB\sin\theta$. A long straight wire carrying a current generates circular magnetic field lines concentric with the wire, meaning the magnetic field vector at any point points tangentially in a plane perpendicular to the wire. Because the electron moves directly towards the wire along the perpendicular radius, its velocity vector is perpendicular to the tangential magnetic field lines, making the angle 90 degrees with a sine value of 1. Substituting the numerical values for charge, velocity, and magnetic field strength results in a force of $1.6 \times 10^{-19} \times 10^7 \times 2 \times 10^{-4}$, which calculates to $3.2 \times 10^{-16} \, \text{N}$.

The acceleration of a charged particle in combined fields is governed by the Lorentz force equation $\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) = m\vec{a}$. When the proton is released from rest, the magnetic force vanishes, meaning the initial westward acceleration is driven entirely by the electric field pointing toward the west with a magnitude of $E = \frac{ma_0}{e}$. When the proton is projected northward, the net westward acceleration increases to $3a_0$, indicating that the magnetic force contributes an additional westward acceleration of $2a_0$. Evaluating the cross product of the northward velocity with this westward magnetic force reveals that the magnetic field must be directed vertically downward with a magnitude of $B = \frac{2ma_0}{ev_0}$.

Any closed current-carrying loop of any shape carrying a current $i$ within a uniform magnetic field experiences a net magnetic force of zero. Because the entire circular coil $ABCD$ constitutes a complete closed path, the vector sum of the forces acting on the segment $AB$ and the remaining segment $BCDA$ must equal zero. Consequently, rearranging the vector relation $\vec{F}_{AB} + \vec{F}_{BCDA} = 0$ demonstrates that the force exerted on the remaining segment is simply $-\vec{F}$.

The magnetic field produced by a circular loop of radius $R$ carrying current $I$ at its centre is given by $B = \frac{\mu_0 I}{2R}$, acting perpendicular to the plane of the loop. Since the two loops lie in mutually perpendicular planes, their respective magnetic fields $B_1 = \frac{\mu_0 I}{2R}$ and $B_2 = \frac{\mu_0 (2I)}{2R}$ are also perpendicular to each other. The resultant magnetic field is therefore obtained by taking the vector sum of these orthogonal components, yielding $B_{\text{net}} = \sqrt{B_1^2 + B_2^2}$. Factoring out the common terms results in $B_{\text{net}} = \frac{\mu_0 I}{2R}\sqrt{1^2 + 2^2}$, which simplifies directly to $\frac{\sqrt{5}\mu_0 I}{2R}$.

A millivoltmeter is converted into an ammeter by connecting a very small resistance, known as a shunt, in parallel with it. The maximum voltage across the instrument is 25 mV and the desired full-scale current is 25 A. By neglecting the small current flowing through the meter itself compared to the total range current, the required shunt resistance is calculated using the formula $S = \frac{V_g}{I}$. Substituting these values gives $\frac{25 \times 10^{-3}}{25}$, which evaluates precisely to $0.001 \, \Omega$.

The operating frequency of the alternating electric field in a cyclotron must equal the cyclotron frequency of the protons, leading to the relation $\nu = \frac{eB}{2\pi m}$, which rearranges to $B = \frac{2\pi m\nu}{e}$. The maximum velocity of the protons at the outer radius $R$ is determined by equating the magnetic force to the centripetal force, giving $v = \frac{eBR}{m} = 2\pi\nu R$. Substituting this velocity expression into the standard kinetic energy formula $K = \frac{1}{2}mv^2$ yields $K = \frac{1}{2}m(2\pi\nu R)^2$, which simplifies to $2m\pi^2\nu^2R^2$.

The radius of curvature for a charged particle moving in a uniform magnetic field is expressed as $R = \frac{\sqrt{2mK}}{Bq}$, where $m$ is mass and $q$ is charge. For a proton, the mass is $m$ and the charge is $e$, whereas an alpha particle has a mass of $4m$ and a charge of $2e$. Substituting these respective values into the radius equation shows that the scaling factors for mass and charge simplify identically for both particles. Since both describe circles of the same radius in the identical magnetic field, their kinetic energies must be equal. Thus, the alpha particle requires a kinetic energy of 1 MeV, matching that of the proton.

A charged particle moving in combined fields experiences the Lorentz force, combining contributions from both electric and magnetic interactions. Because the electron carries a negative charge, the electrostatic force acts in a direction directly opposite to the electric field vector. Meanwhile, the magnetic force vanishes entirely because the velocity vector of the electron and the uniform magnetic field are aligned along the same axis, making the vector cross product zero. Since the net force on the electron points against its direction of motion, it undergoes deceleration, causing its speed to decrease steadily.

A rotating ring carrying a distributed charge constitutes a circular current loop whose magnitude is directly proportional to the charge and the rotational frequency, expressed as $I = qf$. The magnetic induction at the geometric center of any circular current-carrying loop is determined by the Biot-Savart law and simplifies to $\frac{\mu_0 I}{2R}$. Substituting the equivalent current into this standard formula yields $\frac{\mu_0 qf}{2R}$ for the magnetic field at the center.

The magnetic interaction between parallel current-carrying conductors dictates that parallel currents attract while opposing currents repel. Since the magnetic field intensity diminishes with increasing distance from the straight wire, the side of the square loop situated closer to the conductor experiences a stronger magnetic pull than the farther parallel side. Because this closer-range attraction ($F_1$) exceeds the force acting on the distant side ($F_2$), the magnitude inequality $F_1 > F_2$ results in a net attractive force drawing the loop toward the conductor.

Maintaining the main current unchanged in a circuit requires the total equivalent resistance of the galvanometer branch to remain invariant before and after introducing the shunt. Initially, the branch resistance is simply equal to $G$. When the galvanometer is shunted by resistance $S$ and supplemented with a series resistance $R$, the combined resistance of this new arrangement is given by $\frac{GS}{S+G} + R$. Equating this total effective resistance to the original resistance $G$ allows us to isolate $R$ as $G - \frac{GS}{S+G}$, which simplifies directly to $\frac{G^2}{S+G}$.

Any closed current-carrying loop placed in a uniform magnetic field experiences a total magnetic force of zero because the vector sum of forces acting across all its segments must vanish. Given that the uniform magnetic field acts directly along the arm AB, the current flowing through this segment is parallel to the field lines, which means the magnetic force on arm AB is zero. Consequently, setting the vector sum of forces for the triangular loop to zero gives $\vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{AC} = 0$, leading directly to a force of $-\vec{F}$ on the remaining arm AC.

The continuous circular motion of a uniformly distributed charge creates an equivalent electric current determined by the product of the total charge and the frequency of rotation, written as $I = qf$. The magnetic field produced at the center of a circular loop is governed by the relation $\frac{\mu_0 I}{2R}$. Combining these expressions by substituting the current gives a final magnetic induction of $\frac{\mu_0 qf}{2R}$ at the center of the ring.

The net magnetic force on any closed current-carrying loop placed in a uniform magnetic field is always zero. Since the magnetic field acts entirely in the plane ofებელი the square loop, the two arms aligned parallel to the field experience no magnetic force. Consequently, the force on the remaining two arms must balance out the force on the initial arm, meaning the total force exerted on the other three arms combined is the exact negative vector, $-\overset{\rightarrow}{F}$.

Converting a galvanometer into a voltmeter requires connecting a large resistance in series so that the total potential drop across the device matches the desired voltage range without exceeding the current limit. The relationship between the maximum allowable voltage, full-scale deflection current, galvanometer resistance, and the added series resistance is expressed by the equation $V = I_g (G + R)$. Substituting the given values of $30\text{ V}$ for voltage, $30 \times 10^{-3}\text{ A}$ for current, and $100\ \Omega$ for galvanometer resistance yields a total required resistance of $1000\ \Omega$, which leaves a series resistance of $900\ \Omega$ to be added.