Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.
Reason (R): The magnetic field in that region is along the direction of velocity of the electron.
In the light of the above statements, choose the correct answer from the options given below:
A.(A)$(A)$ is false but (R)$(R)$ is true
B.Both (A)$(A)$ and (R)$(R)$ are true and (R)$(R)$ is the correct explanation of (A)$(A)$
C.Both (A)$(A)$ and (R)$(R)$ are true but (R)$(R)$ is NOT the correct explanation of (A)$(A)$
D.(A)$(A)$ is true but (R)$(R)$ is false
Solution & Explanation
### Related Formula
vecF = q(vecv times vecB)$\vec{F} = q(\vec{v} \times \vec{B})$
### Core Logic
For a particle to move with a constant velocity in a straight line inside a magnetic field alone, the net magnetic force must be zero:
vecF = 0 implies vecv parallel vecB$\vec{F} = 0 \implies \vec{v} \parallel \vec{B}$
This means the angle theta$\theta$ between the velocity vector and the magnetic field vector is either 0^circ$0^{\circ}$ or 180^circ$180^{\circ}$. Thus, if the magnetic field is along the direction of velocity, the force is zero, allowing unaccelerated straight-line motion. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.
### Pattern Recognition
Magnetic field cannot change the speed of a charged particle, but it changes direction unless vecv$\vec{v}$ is parallel or anti-parallel to vecB$\vec{B}$, in which case the magnetic force vanishes entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Keywords:#Magnetic force on electron#JEE Main 2025 Evening Q3#Moving Charges and Magnetism JEE Main 2025#Constant velocity in magnetic field
More Moving Charges and Magnetism Previous-Year Questions — Page 2
Q62025Magnetic Force on a Charged Particle
Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge "q" is released at a distance "a" from the wire with a speed mathbfv_0$\mathbf{v}_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance mathbfx$\mathbf{x}$ from the wire. The value of mathbfx$\mathbf{x}$ is [mu_0$[\mu_0$ is vacuum permeability]
### Core Logic
Analyzing motion from path phases mathrmA rightarrow mathrmB$\mathrm{A} \rightarrow \mathrm{B}$ and mathrmB rightarrow mathrmC$\mathrm{B} \rightarrow \mathrm{C}$ inside the coordinate field:
Coordinate trajectory analysis path diagram for Q6vecmathrmB = fracmu_0 mathrmI2pi mathrmr(-hatmathrmk)$\vec{\mathrm{B}} = \frac{\mu_0 \mathrm{I}}{2\pi \mathrm{r}}(-\hat{\mathrm{k}})$
The lorentz magnetic field acceleration rules dictate differential trajectory steps:
int_v_0^0 fracv_x dv_xsqrtv_0^2 - v_x^2 = -fracmu_0 mathrmI q2pi mathrmm int_a^x_1 fracdrr$\int_{v_0}^{0} \frac{v_x dv_x}{\sqrt{v_0^2 - v_x^2}} = -\frac{\mu_0 \mathrm{I q}}{2\pi \mathrm{m}} \int_{a}^{x_1} \frac{dr}{r}$
Solving this integration step gives the position node parameter:
x_1 = a e^-frac2pi m v_0mu_0 I q$x_1 = a e^{-\frac{2\pi m v_0}{\mu_0 I q}}$
Compounding this loop interaction for the turning path phase mathrmB rightarrow mathrmC$\mathrm{B} \rightarrow \mathrm{C}$ gives:
### Step 1: Final Solution Integration
x = x_1 e^-frac2pi m v_0mu_0 I q = a e^-frac4pi m v_0mu_0 I q$x = x_1 e^{-\frac{2\pi m v_0}{\mu_0 I q}} = a e^{-\frac{4\pi m v_0}{\mu_0 I q}}$
Matches criteria for option (4).
### Pattern Recognition
Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work magnetic operations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q212025Motion in a Magnetic Field
A particle of charge 1.6 µC and mass 16 µg is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is ________ s. (Take pi=3.14$\pi=3.14$)
Numerical Answer.Answer: 0 to 0
Solution
### Related Formula
Time Period of circular motion in a magnetic field:
T = frac2pi mqB$T = \frac{2\pi m}{qB}$
### Core Logic
Convert given parameter values into SI baseline metrics:
q = 1.6\ mutextC = 1.6 times 10^-6text C$q = 1.6\ \mu\text{C} = 1.6 \times 10^{-6}\text{ C}$m = 16\ mutextg = 16 times 10^-9text kg$m = 16\ \mu\text{g} = 16 \times 10^{-9}\text{ kg}$B = 6.28text T = 2pitext T$B = 6.28\text{ T} = 2\pi\text{ T}$
### Step 1: Solve for Time Period
T = frac2pi times 16 times 10^-91.6 times 10^-6 times 6.28 = frac6.28 times 16 times 10^-91.6 times 10^-6 times 6.28$T = \frac{2\pi \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28} = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$T = frac16 times 10^-91.6 times 10^-6 = 10 times 10^-3 = 0.01text seconds$T = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} = 10 \times 10^{-3} = 0.01\text{ seconds}$
Rounding to the nearest integer as required for standard integer formatting gives **0**. Circular trajectory of a charged particle in a magnetic field
### Pattern Recognition
Check your prefixes. Micrograms (mutextg$\mu\text{g}$) conversion introduces a factor of 10^-9text kg$10^{-9}\text{ kg}$, not 10^-6$10^{-6}$. If nearest integer is requested, 0.01text s$0.01\text{ s}$ rounds down cleanly to 0.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q52025Ampere's Circuital Law
A long straight wire of a circular cross-section with radius 'a' carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by:
A. Graph (1)
B. Graph (2)
C. Graph (3)
D. Graph (4)
Solution
### Related Formula
B_textin = fracmu_0 I r2pi a^2 implies B_textin propto r$B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2} \implies B_{\text{in}} \propto r$B_textout = fracmu_0 I2pi r implies B_textout propto frac1r $B_{\text{out}} = \frac{\mu_0 I}{2\pi r} \implies B_{\text{out}} \propto \frac{1}{r} $
### Core Logic
Inside the wire (r < a$r < a$), the magnetic field grows linearly with distance r$r$ from the axis. At the surface (r = a$r = a$), it reaches its maximum value B_textmax = fracmu_0 I2pi a$B_{\text{\max}} = \frac{\mu_0 I}{2\pi a}$. Outside the wire (r > a$r > a$), it decays inversely with r$r$. This combination matches the curve shown in Graph (1). magnetic field variation, ampere law plot, current carrying wire
### Pattern Recognition
Solid cylinder current profile: linear inside (B propto r$B \propto r$), hyperbolic outside (B propto 1/r$B \propto 1/r$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q112025Ampere'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
More Moving Charges and Magnetism Questions — jee_main_2025_24_jan_evening
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