Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics) — page 3.
Q21 (2025)
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 $\pi=3.14$)
### Related Formula
Time Period of circular motion in a magnetic field:
$$T = \frac{2\pi m}{qB}$$
### Core Logic
Convert given parameter values into SI baseline metrics:
$$q = 1.6\ \mu\text{C} = 1.6 \times 10^{-6}\text{ C}$$
$$m = 16\ \mu\text{g} = 16 \times 10^{-9}\text{ kg}$$
$$B = 6.28\text{ T} = 2\pi\text{ T}$$
### Step 1: Solve for Time Period
$$T = \frac{2\pi \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28} = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$$
$$T = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} = 10 \times 10^{-3} = 0.01\text{ seconds}$$
Rounding to the nearest integer as required for standard integer formatting gives **0**. {{SOL_IMG1}}
### Pattern Recognition
Check your prefixes. Micrograms ($\mu\text{g}$) conversion introduces a factor of $10^{-9}\text{ kg}$, not $10^{-6}$. If nearest integer is requested, $0.01\text{ s}$ rounds down cleanly to 0.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q3 (2025)
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:
- $(A)$ is false but $(R)$ is true
- Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
- Both $(A)$ and $(R)$ are true but $(R)$ is NOT the correct explanation of $(A)$
- $(A)$ is true but $(R)$ is false
### Related Formula
$$\vec{F} = q(\vec{v} \times \vec{B})$$
### 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:
$$\vec{F} = 0 \implies \vec{v} \parallel \vec{B}$$
This means the angle $\theta$ between the velocity vector and the magnetic field vector is either $0^{\circ}$ or $180^{\circ}$. 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 $\vec{v}$ is parallel or anti-parallel to $\vec{B}$, in which case the magnetic force vanishes entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q5 (2025)
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:
- Graph (1)
- Graph (2)
- Graph (3)
- Graph (4)
### Related Formula
$$B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2} \implies B_{\text{in}} \propto r$$
$$B_{\text{out}} = \frac{\mu_0 I}{2\pi r} \implies B_{\text{out}} \propto \frac{1}{r} $$
### Core Logic
Inside the wire ($r < a$), the magnetic field grows linearly with distance $r$ from the axis. At the surface ($r = a$), it reaches its maximum value $B_{\max} = \frac{\mu_0 I}{2\pi a}$. Outside the wire ($r > a$), it decays inversely with $r$. This combination matches the curve shown in Graph (1). {{SOL_IMG1}}
### Pattern Recognition
Solid cylinder current profile: linear inside ($B \propto r$), hyperbolic outside ($B \propto 1/r$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q11 (2025)
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 $\omega$ as shown in the figure {{IMG1}}. 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, $I_{A} - I_{B}$, for the given Amperian loops is
- $\frac{N^{2}}{2\pi}q\omega$
- $\frac{2\pi}{N}q\omega$
- $\frac{N}{2\pi}q\omega$
- $\frac{N}{\pi}q\omega$
### Related Formula
$$I = \frac{q}{T} = \frac{q\omega}{2\pi}$$
### 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:
$$I_A = \frac{Nq}{\left(\frac{2\pi}{\omega}\right)} = \frac{Nq\omega}{2\pi}$$
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:
$I_B = 0$
Therefore, the difference is:
$$I_A - I_B = \frac{Nq\omega}{2\pi}$${{SOL_IMG1}}
### 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
Q21 (2025)
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 ____. {{IMG1}}
[Take mass of electron $m_e = 9 \times 10^{-31}$ kg, charge of electron $|q_e| = 1.6 \times 10^{-19}$ C, $\mu_0 = 4\pi \times 10^{-7} \frac{N}{A^2}, 1 \text{ ns} = 10^{-9} \text{ s}$]
### Related Formula
Time period of a revolving charge in a magnetic field:
$$T = \frac{2\pi m}{qB}$$
Magnetic field inside a long solenoid:
$B = \mu_0 n I$
### Core Logic
Combining the expressions to isolate $n$ (turns per meter):
$$T = \frac{2\pi m}{q(\mu_0 n I)}$$
Substituting the given constants:
$$75 \times 10^{-9} = \frac{2\pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5}$$
Simplifying terms:
$$75 \times 10^{-9} = \frac{18\pi \times 10^{-31}}{9.6\pi \times 10^{-26} \times n} = \frac{1.875 \times 10^{-5}}{n}$$
$$n = \frac{1.875 \times 10^{-5}}{75 \times 10^{-9}} = 250$$
### Pattern Recognition
The circular motion time period depends exclusively on the field magnitude $B$, completely independent of the orbit's velocity or radius.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
