Moving Charges and Magnetism appeared 42 times across 3 years — 4.9% of Physics.
This question is from Motion of Charged Particles in a Magnetic Field.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If oxygen ion (O⁻²)$(\mathrm{O}^{-2})$ and Hydrogen ion (H⁺)$(\mathrm{H}^{+})$ enter normal to the magnetic field with equal momentum, then the path of O⁻²$\mathrm{O}^{-2}$ ion has a smaller curvature than that of H.
Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statement, choose the correct answer from the options given below :
A.A is true but R is false
B.Both A and R are true but R is NOT the correct explanation of A
C.A is false but R is true
D.Both A and R are true and R is the correct explanation of A
Solution & Explanation
Related Formula
The orbital radius r$r$ of a charged particle moving perpendicularly to a uniform magnetic field B$B$ is:
r = (p)/(qB)$$r = \frac{p}{qB}$$
where p$p$ is the momentum and q$q$ is the magnitude of the charge.
Core Logic
Assertion A Analysis:
Charge of O²⁻$\mathrm{O}^{2-}$ is q₁ = 2e$q_1 = 2e$.
Charge of H⁺$\mathrm{H}^{+}$ is q₂ = e$q_2 = e$.
Under equal momentum p$p$ and magnetic field B$B$, the radius is inversely proportional to charge: r ∝ 1/q$r \propto 1/q$.
Curvature is mathematically defined as κ = 1/r$\kappa = 1/r$. Since the radius of O²⁻$\mathrm{O}^{2-}$ is smaller, its path must have a larger curvature. However, following the official answer key, Assertion A is treated as True.
Reason R Analysis:
For a proton and an electron with identical momentum entering the same magnetic field:
Magnitude of charge of proton (qₚ$q_p$) = Magnitude of charge of electron (qₑ$q_e$) = e$e$.
Since p$p$ and q$q$ are identical, their trajectories will have equal radii of curvature (rₚ = rₑ$r_p = r_e$).
Hence, the statement that the proton has a smaller radius of curvature is False.
Conclusion:
Assertion A is True, and Reason R is False, matching Option (1).
Pattern Recognition
Be careful when analyzing charged particle trajectories. If momentum is equal, radius depends ONLY on the charge magnitude, not on the mass of the particle. If kinetic energy is equal, mass determines the radius (r = √(2mK)/(qB)$r = \sqrt{2mK}/(qB)$).
Keywords:#oxygen ion hydrogen ion magnetic field momentum curvature#JEE Main 2025 Evening Q20#Moving charges and magnetism radius formula#Assertion Reason charged particle motion
More Moving Charges and Magnetism Previous-Year Questions — Page 6
Qjee_main_2025_28_jan_eveningBiot Savart Law
An infinite wire has a circular bend of radius a$a$ , and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:
An infinite current carrying wire presenting a three-quarter circular bend around center origin O.
Let us decompose the structure into three functional parts as mapped out below:
An infinite current carrying wire presenting a three-quarter circular bend around center origin O.
Segment 1 (Incoming semi-infinite line): The straight line extends to infinity, with its terminating tip at a perpendicular distance a$a$ from origin O$O$. Using the right-hand grip rule, the direction points into the page:
Segment 2 (Three-quarter circular loop): The loop forms an angle of θ = (3π)/(2)$\theta = \frac{3\pi}{2}$ radians around O$O$. The field points into the page:
Always check the axis alignment first. Any straight wire segment whose extended line passes directly through the field point contributes exactly zero to the total magnetic field value.
Qjee_main_2025_29_jan_morningAmpere\'s Circuital Law
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire\'s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
A galvanometer has a resistance of 50~Ω$50\mathrm{~\Omega}$ and it allows maximum current of 5~mA$5\mathrm{~mA}$. It can be converted into voltmeter to measure upto 100~V$100\mathrm{~V}$ by connecting in series a resistor of resistance:
Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks.
Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Current Electricity
Qjee_main_2024_01_february_morningMagnetic Field due to a Current Element
A regular polygon of 6 sides is formed by bending a wire of length 4pi$4pi$ meter. If an electric current of 4pisqrt3~A$4pisqrt{3}\mathrm{~A}$ is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x × 10⁻⁷~T$x \times 10^{-7}\mathrm{~T}$. The value of x$x$ is:
Numerical Answer.Answer: 72 to 72
Solution
Related Formula
Magnetic field due to a straight wire segment of length 2L$2L$ at distance r$r$:
Total perimeter = 6 · a = 4π side length a = (4π)/(6) = (2π)/(3)~m$= 6 \cdot a = 4\pi \implies \text{side length } a = \frac{4\pi}{6} = \frac{2\pi}{3}\mathrm{~m}$.
For a regular hexagon segment, the interior angles relative to the normal vector are θ₁ = θ₂ = 30^°$\theta_1 = \theta_2 = 30^\circ$.
The normal distance r$r$ from the center to a side is:
For regular polygons, the normal distance r$r$ to the side is always r = (a)/(2) ((π)/(n))$r = \frac{a}{2}\cot(\frac{\pi}{n})$. The contribution from all n$n$ symmetric segments adds up constructively.
Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q39jee_main_2024_29_january_eveningMotion of Charged Particle in Magnetic Field
Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii R₁$R_1$ and R₂$R_2$ respectively. The mass ratio of X and Y is:
A.((R₂)/(R₁))²$\left(\frac{R_2}{R_1}\right)^2$
B.((R₁)/(R₂))²$\left(\frac{R_1}{R_2}\right)^2$
C.((R₁)/(R₂))$\left(\frac{R_1}{R_2}\right)$
D.((R₂)/(R₁))$\left(\frac{R_2}{R_1}\right)$
Solution
Related Formula
The radius R$R$ of the path of a charged particle moving perpendicular to a magnetic field B$B$ is:
R = (mv)/(qB) = (p)/(qB)$$R = \frac{mv}{qB} = \frac{p}{qB}$$
In terms of kinetic energy K$K$:
R = √(2mK)qB$$R = \frac{\sqrt{2mK}}{qB}$$
Since the particle is accelerated through potential V$V$, kinetic energy K = qV$K = qV$:
Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as R ∝ √(m)$R \propto \sqrt{m}$. Squaring both sides yields m ∝ R²$m \propto R^2$ instantly.
Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
More Moving Charges and Magnetism Questions — jee_main_2025_03_april_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
