Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics).
Q2 (2025)
In a moving coil galvanometer, two moving coils $M_1$ and $M_2$ have the following particulars:
$$\begin{aligned} R_1 &= 5 \ \Omega, \quad N_1 = 15, \quad A_1 = 3.6 \times 10^{-3} \ \mathrm{m}^2, \quad B_1 = 0.25 \ \mathrm{T} \\ R_2 &= 7 \ \Omega, \quad N_2 = 21, \quad A_2 = 1.8 \times 10^{-3} \ \mathrm{m}^2, \quad B_2 = 0.50 \ \mathrm{T} \end{aligned}$$
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$ ?
- $1:1$
- $1:4$
- $1:3$
- $1:2$
### Related Formula
$$\text{Voltage Sensitivity } (V_s) = \frac{\theta}{V} = \frac{N B A}{C R}$$
where:
$N$ = number of turns
$B$ = magnetic field
$A$ = area of the coil
$C$ = torsional constant of the spring
$R$ = resistance of the coil
### Core Logic
Since the torsional constant $C$ is the same for both coils, the ratio of voltage sensitivities of $M_1$ and $M_2$ is:
$$\frac{(V_s)_1}{(V_s)_2} = \left(\frac{N_1 A_1 B_1}{N_2 A_2 B_2}\right) \cdot \left(\frac{R_2}{R_1}\right)$$
We are given the following values:
- Coil 1: $R_1 = 5 \ \Omega$, $N_1 = 15$, $A_1 = 3.6 \times 10^{-3} \ \mathrm{m}^2$, $B_1 = 0.25 \ \mathrm{T}$
- Coil 2: $R_2 = 7 \ \Omega$, $N_2 = 21$, $A_2 = 1.8 \times 10^{-3} \ \mathrm{m}^2$, $B_2 = 0.50 \ \mathrm{T}$
### Step 1: Calculate the ratio
Substitute the values into the formula:
$$\frac{(V_s)_1}{(V_s)_2} = \left(\frac{15 \times 3.6 \times 10^{-3} \times 0.25}{21 \times 1.8 \times 10^{-3} \times 0.50}\right) \times \frac{7}{5}$$
Simplify the terms within the brackets:
- $\frac{3.6 \times 10^{-3}}{1.8 \times 10^{-3}} = 2$
- $\frac{0.25}{0.50} = \frac{1}{2}$
$$\frac{15 \times 2 \times \frac{1}{2}}{21} = \frac{15}{21} = \frac{5}{7}$$
Multiplying by $\frac{R_2}{R_1} = \frac{7}{5}$:
$$\frac{(V_s)_1}{(V_s)_2} = \frac{5}{7} \times \frac{7}{5} = 1$$
Thus, the ratio is $1:1$.
### Pattern Recognition
Sees: Galvanometer sensitivity comparison with different parameters.
Trap: Confusing Current Sensitivity with Voltage Sensitivity. Current sensitivity is $\frac{NBA}{C}$ (independent of $R$), while voltage sensitivity is $\frac{NBA}{C R}$ (depends on $R$).
Shortcut: Write the ratio as $\frac{(I_s)_1}{(I_s)_2} \times \frac{R_2}{R_1}$ to keep calculations clean.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q89 (2025)
Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of magnetic field at an axial distance '$x$' from the center. For $x:R = 3:4$, $\frac{B_2}{B_1}$ is:
- 4:5
- 16:25
- 64:125
- 25:16
### Related Formula
$$B_1 = \frac{\mu_0 I}{2R}$$
$$B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
$$B_2 = B_1 \sin^3\theta \quad \text{where } \sin\theta = \frac{R}{\sqrt{R^2 + x^2}}$$
### Core Logic
Given the ratio $x : R = 3 : 4$. Let $x = 3k$ and $R = 4k$.
The distance to the element is:
$$\sqrt{R^2 + x^2} = \sqrt{(4k)^2 + (3k)^2} = 5k$$
The sine of the semi-vertical angle subtended at the axial point is:
$$\sin\theta = \frac{R}{\sqrt{R^2 + x^2}} = \frac{4k}{5k} = \frac{4}{5}$$
We know that the axial magnetic field relates to the central magnetic field as follows:
$$\frac{B_2}{B_1} = \sin^3\theta = \left(\frac{4}{5}\right)^3 = \frac{64}{125}$$
### Step 1: Final Conclusion
The ratio $\frac{B_2}{B_1}$ is $64:125$.
### Pattern Recognition
For axial magnetic field problems, bypass manual calculation of $(R^2+x^2)^{3/2}$ by writing the relation directly in terms of the subtended angle: $B_{\text{axial}} = B_{\text{center}} \sin^3\theta$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Magnetic Effects of Current
Q20 (2025)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If oxygen ion $(\mathrm{O}^{-2})$ and Hydrogen ion $(\mathrm{H}^{+})$ enter normal to the magnetic field with equal momentum, then the path of $\mathrm{O}^{-2}$ 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 is true but R is false
- Both A and R are true but R is NOT the correct explanation of A
- A is false but R is true
- Both A and R are true and R is the correct explanation of A
### Related Formula
The orbital radius $r$ of a charged particle moving perpendicularly to a uniform magnetic field $B$ is:
$$r = \frac{p}{qB}$$
where $p$ is the momentum and $q$ is the magnitude of the charge.
### Core Logic
Assertion A Analysis:
- Charge of $\mathrm{O}^{2-}$ is $q_1 = 2e$.
- Charge of $\mathrm{H}^{+}$ is $q_2 = e$.
- Under equal momentum $p$ and magnetic field $B$, the radius is inversely proportional to charge: $r \propto 1/q$.
- Therefore, $r_{\mathrm{O}^{2-}} = \frac{r_{\mathrm{H}^{+}}}{2}$.
- Curvature is mathematically defined as $\kappa = 1/r$. Since the radius of $\mathrm{O}^{2-}$ 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 ($q_p$) = Magnitude of charge of electron ($q_e$) = $e$.
- Since $p$ and $q$ are identical, their trajectories will have equal radii of curvature ($r_p = r_e$).
- 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 ($r = \sqrt{2mK}/(qB)$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q229 (2025)
Uniform magnetic fields of different strengths $(\mathrm{B}_1$ and $\mathrm{B}_2)$ , both normal to the plane of the paper exist as shown in the figure. A charged particle of mass $m$ and charge $q$ , at the interface at an instant, moves into the region 2 with velocity $v$ and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?
{{IMG1}}
(Consider the velocity of the particle to be normal to the magnetic field and $\mathrm{B}_2 > \mathrm{B}_1$ )
- $\frac{\mathrm{mv}}{\mathrm{qB}_1}\left(1 - \frac{\mathrm{B}_2}{\mathrm{B}_1}\right)\times 2$
- $\frac{\mathrm{mv}}{\mathrm{qB}_1}\left(1 - \frac{\mathrm{B}_1}{\mathrm{B}_2}\right)$
- $\frac{\mathrm{mv}}{\mathrm{qB}_1}\left(1 - \frac{\mathrm{B}_2}{\mathrm{B}_1}\right)$
- $\frac{\mathrm{mv}}{\mathrm{qB}_1}\left(1 - \frac{\mathrm{B}_1}{\mathrm{B}_2}\right)\times 2$
### Related Formula
The radius of a circular trajectory of a charged particle in a uniform magnetic field perpendicular to its velocity is:
$$R = \frac{mv}{qB}$$
### Core Logic
$\text{Starting point} \rightarrow \text{A}$
$\text{Ending point} \rightarrow \text{C}$
$\therefore \text{Net displacement} = \text{AC}$
$\text{AC} = \text{CD} - \text{AD}$
$\text{AC} = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}$
$\text{AC} = \frac{2mv}{qB_1} \left[ 1 - \frac{B_1}{B_2} \right]$
### Pattern Recognition
Sees: Particle traversing two different perpendicular fields across an interface.
Shortcut: The displacement on completing a loop across a boundary between two fields always equals $2(R_{\text{large}} - R_{\text{small}})$. Factoring out the term with $B_1$ in the denominator leaves the factor $\left(1 - B_1/B_2\right)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q242 (2025)
A particle of charge $q$ , mass $m$ and kinetic energy $E$ enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius (r). Which of the following curves represents the variation of $r$ with $E$ ?
### Related Formula
The magnetic force provides the centripetal force for circular motion:
$$\frac{mv^2}{r} = qvB \implies r = \frac{mv}{qB}$$
Kinetic energy $E$ is related to momentum $p = mv$ by:
$p = \sqrt{2mE}$
### Core Logic
Express the radius $r$ in terms of kinetic energy $E$:
$$r = \frac{\sqrt{2mE}}{qB}$$
Since $m$, $q$, and $B$ are constants:
$$r \propto \sqrt{E} \implies r^2 \propto E$$
### Step 1: Graph Identification
The relation $r \propto \sqrt{E}$ describes a parabola that opens towards the horizontal energy axis (concave down, starting at origin $(0,0)$).
{{SOL_IMG1}}
Reviewing the options:
- Curve 1 (linear graph) - Incorrect
- Curve 2 (parabola opening vertically) - Incorrect
- Curve 3 (hyperbola / decaying curve) - Incorrect
- Curve 4 (square root shape / parabola opening horizontally) - **Correct**
### Pattern Recognition
Sees: Circular radius $r$ versus kinetic energy $E$ graph.
Shortcut: Radius is proportional to momentum, which grows as the square root of kinetic energy ($r \propto \sqrt{E}$). Any $y \propto \sqrt{x}$ plot is a sideways-opening parabola starting at $(0,0)$ with decreasing slope.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
