Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics) — page 8.
Q29 (2026)
The current passing through a conducting loop in the form of equilateral triangle of side $4\sqrt{3} \, \mathrm{cm}$ is 2A. The magnetic field at its centroid is $\alpha \times 10^{-5}\mathrm{T}$ . The value of $\alpha$ is ____.
(Given: $\mu_{\mathrm{o}} = 4\pi \times 10^{-7}$ SI units)
- $2\sqrt{3}$
- $\sqrt{3}$
- $3\sqrt{3}$
- $\frac{\sqrt{3}}{2}$
### Related Formula
$$B = \frac{\mu_0 I}{4\pi d} [\sin \theta_1 + \sin \theta_2]$$
### Core Logic
{{SOL_IMG1}}
For an equilateral triangle, the perpendicular distance $d$ from centroid to any side is given by $d = \frac{a}{2\sqrt{3}}$, where $a = 4\sqrt{3} \, \mathrm{cm}$.
$$d = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \, \mathrm{cm} = 2 \times 10^{-2} \, \mathrm{m}$$
The angles subtended by the side at the centroid are $\theta_1 = 60^{\circ}$ and $\theta_2 = 60^{\circ}$.
### Step 1: Field due to one side
$$B_1 = \frac{\mu_0}{4\pi} \cdot \frac{I}{d} (\sin 60^{\circ} + \sin 60^{\circ})$$
$$B_1 = 10^{-7} \times \frac{2}{2 \times 10^{-2}} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right)$$
$$B_1 = 10^{-5} \times \sqrt{3} \, \mathrm{T}$$
### Step 2: Total Magnetic Field
Since there are 3 identical sides and their field vectors point in the same direction at the centroid:
$$B_{\text{net}} = 3 \times B_1 = 3 \times \sqrt{3} \times 10^{-5} \, \mathrm{T}$$
Comparing with $\alpha \times 10^{-5}\mathrm{T}$, we get $\alpha = 3\sqrt{3}$.
### Pattern Recognition
For regular polygons of $n$ sides, $B_{\text{centroid}} = n \cdot B_{\text{side}}$. An equilateral triangle has $n=3$, $d = a/(2\sqrt{3})$, and angles are always $60^{\circ}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q26 (2026)
Match the List-I with List-II
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I</th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Magnetic induction</td><td style="border: 1px solid #888; padding: 8px;">I. $ML\ T^{-2}A^{-2}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Magnetic flux</td><td style="border: 1px solid #888; padding: 8px;">II. $ML^{2}\ T^{-2}A^{-2}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Magnetic permeability</td><td style="border: 1px solid #888; padding: 8px;">III. $ML^{0}\ T^{-2}A^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Self inductance</td><td style="border: 1px solid #888; padding: 8px;">IV. $ML^{2}\ T^{-2}A^{-1}$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{A-IV, B-III, C-I, D-II}$
- $\text{A-III, B-IV, C-II, D-I}$
- $\text{A-I, B-III, C-IV, D-II}$
- $\text{A-III, B-IV, C-I, D-II}$
### Related Formula
$F = qvB$
$$\phi = B \cdot \text{Area}$$
$$U = \frac{1}{2} L I^{2}$$
### Core Logic
For Magnetic induction (B):
$$[B] = \left[ \frac{F}{qv} \right] = [MT^{-2}A^{-1}]$$
So, A matches III.
For Magnetic Flux ($\phi$):
$$[\phi] = [B] \cdot [\text{Area}] = [ML^{2}T^{-2}A^{-1}]$$
So, B matches IV.
For Magnetic Permeability ($\mu$):
$$[\mu] = [MLT^{-2}A^{-2}]$$
So, C matches I.
For Self inductance (L):
Using $U = \frac{1}{2} LI^{2}$,
$$[L] = [ML^{2}T^{-2}A^{-1}]$$
Wait, the given option II is $[ML^2 T^{-2} A^{-2}]$.
Let's re-verify: Energy $U = [ML^2 T^{-2}]$. $I^2 = [A^2]$. So $L = [ML^2 T^{-2} A^{-2}]$.
So, D matches II.
{{SOL_IMG1}}
### Step 1: Final Conclusion
A-III, B-IV, C-I, D-II. Option (4) is correct.
### Pattern Recognition
Dimensional analysis of electromagnetic quantities frequently hinges on knowing formulas for force, flux, and energy. Deriving from $F=qvB$ and $U=\frac{1}{2}LI^2$ is the fastest approach.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Electromagnetic Induction
Q32 (2026)
{{IMG1}}
Two identical circular loops P and Q each of radius r are lying in parallel planes such that they have common axis. The current through P and Q are I and 4I respectively in clockwise direction as seen from O. The net magnetic field at O is:
- $\frac{3\mu_{\mathrm{o}}\mathrm{I}}{4\sqrt{2}\mathrm{r}} \text{ toward P}$
- $\frac{\mu_{\mathrm{o}}\mathrm{I}}{4\sqrt{2}\mathrm{r}} \text{ toward P}$
- $\frac{\mu_{\mathrm{o}}\mathrm{I}}{4\sqrt{2}\mathrm{r}} \text{ towards Q}$
- $\frac{3\mu_{\mathrm{o}}\mathrm{I}}{4\sqrt{2}\mathrm{r}} \text{ towards Q}$
### Related Formula
$$B = \frac{\mu_0 i R^2}{2(x^2 + R^2)^{3/2}}$$
### Core Logic
The net magnetic field at O is the vector sum of fields from both loops.
Since the currents are in the same relative orientation (clockwise from O), their magnetic fields at O will point in opposite directions.
$$B_{\text{net}} = B_1 - B_2$$
{{SOL_IMG1}}
### Step 1: Superposition of Fields
Magnetic field due to loop Q (carrying 4I) towards Q, and due to loop P (carrying I) towards P.
$$B_{\text{net}} = \frac{\mu_0 (4\mathrm{i}) R^2}{2(R^2 + R^2)^{3/2}} - \frac{\mu_0 (\mathrm{i}) R^2}{2(R^2 + R^2)^{3/2}}$$
$$B_{\text{net}} = \frac{3\mu_0 \mathrm{i} R^2}{2(2R^2)^{3/2}}$$
### Step 2: Final Calculation
$$B_{\text{net}} = \frac{3\mu_0 \mathrm{i} R^2}{2(2\sqrt{2}R^3)} = \frac{3\mu_0 \mathrm{i}}{4\sqrt{2}R}$$
The direction is towards Q because the field from the loop carrying $4I$ is dominant.
### Pattern Recognition
When symmetrical coils carry opposing fields along their axis at equidistance, you simply subtract their current multipliers (4I - I = 3I) and apply the standard axial magnetic field formula once.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q38 (2026)
A moving coil galvanometer of resistance 100 Ω shows a full scale deflection for a current of 1 mA. The value of resistance required to convert this galvanometer into an ammeter, showing full scale deflection for a current of 5 mA, is ____ Ω
- $25$
- $10$
- $0.5$
- $2.5$
### Related Formula
$$r_s = \frac{i_g \times G}{i - i_g} = \frac{G}{\left(\frac{i}{i_g} - 1\right)}$$
### Core Logic
{{SOL_IMG1}}
Given:
Galvanometer resistance, $G = 100 \, \Omega$
Full scale deflection current, $i_g = 1 \text{ mA}$
Target current, $i = 5 \text{ mA}$
### Step 1: Calculate Shunt Resistance
Using the shunt resistance formula:
$$r_s = \frac{100}{\left(\frac{5}{1} - 1\right)}$$
$$r_s = \frac{100}{4} = 25 \, \Omega$$
### Pattern Recognition
When upgrading range $n$ times ($n = i / i_g$), the required parallel shunt is simply $G / (n - 1)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q34 (2026)
Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is ____.
{{IMG1}}
$(\mu_{0} = 4\pi \times 10^{-7}\mathrm{T\cdot m / A})$
- $6 \times 10^{-7} \mathrm{~N} \text{ towards P}$
- $6 \times 10^{-6} \mathrm{~N} \text{ towards R}$
- $6 \times 10^{-7} \mathrm{~N} \text{ towards R}$
- $6 \times 10^{-6} \mathrm{~N} \text{ towards P}$
### Related Formula
$$F = \frac{\mu_0 I_1 I_2 \ell}{2\pi d}$$
### Core Logic
Wires with current in opposite directions repel each other. Wires with current in the same direction attract each other. Calculate the individual force vectors on wire Q due to wires P and R.
### Step 1: Force due to Wire P
Wire P (3A up) and Wire Q (1A down) carry current in opposite directions. So, P repels Q towards the right (towards R).
$$F_{PQ} = \frac{\mu_0 I_P I_Q \ell}{2\pi d_1} = \frac{\mu_0 (3)(1) \ell}{2\pi (3 \times 10^{-2})}$$
### Step 2: Force due to Wire R
Wire R (2A down) and Wire Q (1A down) carry current in the same direction. So, R attracts Q towards the right (towards R).
$$F_{RQ} = \frac{\mu_0 I_R I_Q \ell}{2\pi d_2} = \frac{\mu_0 (2)(1) \ell}{2\pi (2 \times 10^{-2})}$$
### Step 3: Net Force Calculation
Both forces act in the same direction (towards R).
$$F_{\text{net}} = F_{PQ} + F_{RQ} = \frac{\mu_0}{2\pi} I_Q \left( \frac{I_P}{d_1} + \frac{I_R}{d_2} \right) \ell$$
$$F_{\text{net}} = 2 \times 10^{-7} \times 1 \times \left( \frac{3}{3 \times 10^{-2}} + \frac{2}{2 \times 10^{-2}} \right) \times 15 \times 10^{-2}$$
$$= 2 \times 10^{-7} \times \left( \frac{1}{10^{-2}} + \frac{1}{10^{-2}} \right) \times 15 \times 10^{-2}$$
$$= 2 \times 10^{-7} \times (200) \times 0.15$$
$$= 400 \times 10^{-7} \times 0.15 = 60 \times 10^{-7} = 6 \times 10^{-6} \mathrm{~N}$$
### Step 4: Direction
The net force is $6 \times 10^{-6} \mathrm{~N}$ towards R.
### Pattern Recognition
Always map the direction (attract/repel) before plugging in numbers. Same direction = Attract. Opposite direction = Repel.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
