Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If oxygen ion (mathrmO^-2)$(\mathrm{O}^{-2})$ and Hydrogen ion (mathrmH^+)$(\mathrm{H}^{+})$ enter normal to the magnetic field with equal momentum, then the path of mathrmO^-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 = fracpqB$$r = \frac{p}{qB}$$
where p$p$ is the momentum and q$q$ is the magnitude of the charge.
### Core Logic
Assertion A Analysis:
- Charge of mathrmO^2-$\mathrm{O}^{2-}$ is q_1 = 2e$q_1 = 2e$.
- Charge of mathrmH^+$\mathrm{H}^{+}$ is q_2 = e$q_2 = e$.
- Under equal momentum p$p$ and magnetic field B$B$, the radius is inversely proportional to charge: r propto 1/q$r \propto 1/q$.
- Therefore, r_mathrmO^2- = fracr_mathrmH^+2$r_{\mathrm{O}^{2-}} = \frac{r_{\mathrm{H}^{+}}}{2}$.
- Curvature is mathematically defined as kappa = 1/r$\kappa = 1/r$. Since the radius of mathrmO^2-$\mathrm{O}^{2-}$ 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 (q_p$q_p$) = Magnitude of charge of electron (q_e$q_e$) = e$e$.
- Since p$p$ and q$q$ are identical, their trajectories will have equal radii of curvature (r_p = r_e$r_p = r_e$).
- 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 = sqrt2mK/(qB)$r = \sqrt{2mK}/(qB)$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
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 3
Q11jee_main_2025_24_jan_eveningAmpere's Circuital Law
N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity omega$\omega$ as shown in the figure The figure illustrates a rotating ring of charges with two distinct Amperian paths A and B intersecting the ring path.. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, I_A - I_B$I_{A} - I_{B}$, for the given Amperian loops is
A.fracN^22piqomega$\frac{N^{2}}{2\pi}q\omega$
B.frac2piNqomega$\frac{2\pi}{N}q\omega$
C.fracN2piqomega$\frac{N}{2\pi}q\omega$
D.fracNpiqomega$\frac{N}{\pi}q\omega$
Solution
### Related Formula
I = fracqT = fracqomega2pi$$I = \frac{q}{T} = \frac{q\omega}{2\pi}$$
### Core Logic
The loop A encloses one of the moving point charges as it moves past, giving a current contribution localized to that cross-sectional segment intersection:
I_A = fracNqleft(frac2piomega
ight) = fracNqomega2pi$$I_A = \frac{Nq}{\left(\frac{2\pi}{\omega}
ight)} = \frac{Nq\omega}{2\pi}$$
Loop B encloses the entire loop surface coplanar or enclosing the ring structure fully without clipping individual passing current tracks perpendicularly in the same directional fashion, resulting in zero net cross-surface passing enclosed current:
I_B = 0$I_B = 0$
Therefore, the difference is:
I_A - I_B = fracNqomega2pi$$I_A - I_B = \frac{Nq\omega}{2\pi}$$The figure illustrates a rotating ring of charges with two distinct Amperian paths A and B intersecting the ring path.
### Pattern Recognition
A current loop has net passing current across a large overarching bounding box equal to zero if it doesn't cross the boundary surfaces symmetrically.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q21jee_main_2025_24_jan_eveningSolenoid
A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75ns. The number of turns per metre in the solenoid is ____. The figure details a thick solenoid cylinder with a internal cross section displaying a charge tracking loop.
[Take mass of electron m_e = 9 times 10^-31$m_e = 9 \times 10^{-31}$ kg, charge of electron |q_e| = 1.6 times 10^-19$|q_e| = 1.6 \times 10^{-19}$ C, mu_0 = 4pi times 10^-7 fracNA^2, 1 text ns = 10^-9 text s$\mu_0 = 4\pi \times 10^{-7} \frac{N}{A^2}, 1 \text{ ns} = 10^{-9} \text{ s}$]
Numerical Answer.Answer: 250 to 250
Solution
### Related Formula
Time period of a revolving charge in a magnetic field:
T = frac2pi mqB$$T = \frac{2\pi m}{qB}$$
Magnetic field inside a long solenoid:
B = mu_0 n I$B = \mu_0 n I$
### Core Logic
Combining the expressions to isolate n$n$ (turns per meter):
T = frac2pi mq(mu_0 n I)$$T = \frac{2\pi m}{q(\mu_0 n I)}$$
Substituting the given constants:
75 times 10^-9 = frac2pi times 9 times 10^-311.6 times 10^-19 times 4pi times 10^-7 times n times 1.5$$75 \times 10^{-9} = \frac{2\pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5}$$
Simplifying terms:
75 times 10^-9 = frac18pi times 10^-319.6pi times 10^-26 times n = frac1.875 times 10^-5n$$75 \times 10^{-9} = \frac{18\pi \times 10^{-31}}{9.6\pi \times 10^{-26} \times n} = \frac{1.875 \times 10^{-5}}{n}$$n = frac1.875 times 10^-575 times 10^-9 = 250$$n = \frac{1.875 \times 10^{-5}}{75 \times 10^{-9}} = 250$$
### Pattern Recognition
The circular motion time period depends exclusively on the field magnitude B$B$, completely independent of the orbit's velocity or radius.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q24jee_main_2025_24_jan_morningMagnetic Field due to a Current Element
A current of 5A exists in a square loop of side frac1sqrt2text m$\frac{1}{\sqrt{2}}\text{ m}$ Then the magnitude of the magnetic field B at the centre of the square loop will be ptimes10^-6text T$p\times10^{-6}\text{ T}$ where, value of p is [Take mu_0=4pitimes10^-7text T mA^-1$\mu_{0}=4\pi\times10^{-7}\text{ T mA}^{-1}$].
Numerical Answer.Answer: 8 to 8
Solution
### Related Formula
The magnetic field B_1$B_{1}$ produced by a straight wire segment carrying current I$I$ at a perpendicular distance d$d$ is given by the Biot-Savart relation:
B_1 = fracmu_0I4pi d(sintheta_1 + sintheta_2)$$B_{1} = \frac{\mu_{0}I}{4\pi d}(\sin\theta_{1} + \sin\theta_{2})$$
### Core Logic
As shown in the square geometric layout Magnetic Field due to a Current Element diagram for Q24 - JEE Main 2025 Morning, the perpendicular distance from any side to the central origin point is exactly half the total side length :
d = fraca2 = frac12sqrt2text m$$d = \frac{a}{2} = \frac{1}{2\sqrt{2}}\text{ m}$$
Connecting the ends of a side to the center forms internal angles of theta_1 = theta_2 = 45^circ$\theta_{1} = \theta_{2} = 45^{\circ}$.
### Step 1: Summing the Contributions
Calculate the magnetic field contribution from a single side :
B_1 = frac10^-7 times 5frac12sqrt2 left(sin 45^circ + sin 45^circ
ight) = 10^-7 times 10sqrt2 times left(frac2sqrt2
ight) = 2 times 10^-6text T$$B_{1} = \frac{10^{-7} \times 5}{\frac{1}{2\sqrt{2}}} \left(\sin 45^{\circ} + \sin 45^{\circ}
ight) = 10^{-7} \times 10\sqrt{2} \times \left(\frac{2}{\sqrt{2}}
ight) = 2 \times 10^{-6}\text{ T}$$
Since the current flows in the same rotational direction along all four sides, their individual magnetic fields add constructively at the center :
B_textnet = 4 times B_1 = 4 times (2 times 10^-6text T) = 8 times 10^-6text T$$B_{\text{net}} = 4 \times B_{1} = 4 \times (2 \times 10^{-6}\text{ T}) = 8 \times 10^{-6}\text{ T}$$
Comparing this with p times 10^-6text T$p \times 10^{-6}\text{ T}$ , we get:
p = 8$p = 8$
### Pattern Recognition
The magnetic field at the center of any square loop simplifies to the standard formula: B = frac2sqrt2mu_0Ipi a$B = \frac{2\sqrt{2}\mu_{0}I}{\pi a}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
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 50mathrm~Omega$50\mathrm{~\Omega}$ and it allows maximum current of 5mathrm~mA$5\mathrm{~mA}$. It can be converted into voltmeter to measure upto 100mathrm~V$100\mathrm{~V}$ by connecting in series a resistor of resistance:
A.5975mathrm~Omega$5975\mathrm{~\Omega}$
B.20050mathrm~Omega$20050\mathrm{~\Omega}$
C.19950mathrm~Omega$19950\mathrm{~\Omega}$
D.19500mathrm~Omega$19500\mathrm{~\Omega}$
Solution
### Related Formula
Voltmeter series conversion formula:
V = I_g(R_g + R)$$V = I_g(R_g + R)$$R = fracVI_g - R_g$$R = \frac{V}{I_g} - R_g$$
### Core Logic
Given data:
R_g = 50mathrm~Omega$R_g = 50\mathrm{~\Omega}$, I_g = 5mathrm~mA = 5 times 10^-3mathrm~A$I_g = 5\mathrm{~mA} = 5 \times 10^{-3}\mathrm{~A}$, target voltage range V = 100mathrm~V$V = 100\mathrm{~V}$.
Substitute values:
R = frac1005 times 10^-3 - 50$$R = \frac{100}{5 \times 10^{-3}} - 50$$
### Step 1: Complete Arithmetic Evaluation
R = 20000 - 50 = 19950mathrm~Omega$R = 20000 - 50 = 19950\mathrm{~\Omega}$
### Pattern Recognition
Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Current Electricity
More Moving Charges and Magnetism Questions — jee_main_2025_03_april_evening
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