Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics) — page 2.
Q14 (2025)
Figure shows a current carrying square loop ABCD of edge length is 'a' lying in a plane. If the resistance of the ABC part is $r$ and that of ADC part is $2r$, then the magnitude of the resultant magnetic field at centre of the square loop is:
{{IMG1}}
- $\frac{3\pi\mu_{\mathrm{o}} I}{\sqrt{2}a}$
- $\frac{\mu_0 I}{2\pi a}$
- $\frac{\sqrt{2}\mu_{\mathrm{o}} I}{3\pi a}$
- $\frac{2\mu_{\mathrm{o}} I}{3\pi a}$
### Related Formula
$$B_{\text{wire}} = \frac{\mu_0 I}{4\pi d} \left(\sin\theta_1 + \sin\theta_2\right)$$
where,
$B_{\text{wire}}$ = magnetic field due to a straight wire segment
$d$ = perpendicular distance from center to wire segment ($d = a/2$)
$\theta_1, \theta_2$ = subtended angles at the center
### Core Logic
The loop parts ABC and ADC are in parallel. Let total current entering corner A and leaving corner C be $I$.
- Resistance of path ABC = $r$
- Resistance of path ADC = $2r$
Using current divider rule:
- Current in path ABC, $I_1 = \frac{2r}{r + 2r} I = \frac{2}{3} I$
- Current in path ADC, $I_2 = \frac{r}{r + 2r} I = \frac{1}{3} I$
Now, calculate the magnetic field contribution at center $O$:
- Distance of center from each of the 4 wire segments is $d = \frac{a}{2}$.
- For each segment, the angles are $\theta_1 = \theta_2 = 45^\circ$:
$$B_{\text{segment}} = \frac{\mu_0 i}{4\pi (a/2)} \left(\sin 45^\circ + \sin 45^\circ\right) = \frac{\mu_0 i}{2\pi a} \sqrt{2}$$
### Step 1: Summing the Fields at the Center
Using right-hand rule to find the direction of magnetic fields:
- Paths AB and BC (carrying $I_1$ clockwise) create fields pointing into the page ($-\hat{k}$).
- Paths AD and DC (carrying $I_2$ counter-clockwise) create fields pointing out of the page ($+\hat{k}$).
$$B_{\text{net}} = 2 B_{\text{segment}}(I_2) - 2 B_{\text{segment}}(I_1)$$
$$B_{\text{net}} = 2 \left[ \frac{\mu_0 (I/3)}{2\pi a} \sqrt{2} \right] - 2 \left[ \frac{\mu_0 (2I/3)}{2\pi a} \sqrt{2} \right]$$
$$B_{\text{net}} = \frac{\sqrt{2}\mu_0 I}{\pi a} \left( \frac{1}{3} - \frac{2}{3} \right) = -\frac{\sqrt{2}\mu_0 I}{3\pi a}\hat{k}$$
Taking the magnitude of the field:
$$|B_{\text{net}}| = \frac{\sqrt{2}\mu_{\mathrm{o}} I}{3\pi a}$$
### Pattern Recognition
Sees: Current divides into parallel paths of a symmetric loop.
Trap: In a completely symmetric square loop (equal resistances), the net magnetic field at the center is exactly $0$. However, here the paths have unequal resistances ($r$ and $2r$), which prevents total cancellation.
Shortcut: Find the net effective current imbalance $\Delta I = I_1 - I_2 = \frac{2}{3}I - \frac{1}{3}I = \frac{1}{3}I$. The magnitude is simply the magnetic field contribution of two sides carrying this net imbalance. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Magnetic Effects of Current
Q21 (2025)
The magnetic field inside a 200 turns solenoid of radius $10\mathrm{~cm}$ is $2.9 \times 10^{-4}$ Tesla. If the solenoid carries a current of $0.29\mathrm{~A}$ , then the length of the solenoid is ______ $\pi \mathrm{~cm}$.
### Related Formula
$$B = \mu_0 n I = \mu_0 \left(\frac{N}{\ell}\right) I$$
where,
$B$ = magnetic field inside the solenoid
$N$ = total number of turns
$\ell$ = length of the solenoid
$I$ = current inside the wire
### Core Logic
Assuming a long standard solenoid, express the formula to solve for length $\ell$:
$$\ell = \frac{\mu_0 N I}{B}$$
Substitute the given numeric parameters:
$$\ell = \frac{(4\pi \times 10^{-7}) \times 200 \times 0.29}{2.9 \times 10^{-4}}$$
$$\ell = \frac{4\pi \times 10^{-7} \times 200 \times 0.29}{29 \times 10^{-5}} = 8\pi \times 10^{-2} \mathrm{~m} = 8\pi \mathrm{~cm}$$
Since the target unit suffix is specified as $\pi \mathrm{~cm}$, the missing integer coefficient is exactly 8.
### Pattern Recognition
Radius ($10\mathrm{~cm}$) serves as dummy unneeded information for the idealized long solenoid expression. Always look at the required final unit structure to avoid simple metric scalar parsing errors.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q6 (2025)
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 $\mathbf{v}_0$ 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 $\mathbf{x}$ from the wire. The value of $\mathbf{x}$ is $[\mu_0$ is vacuum permeability]
- $a\left[1 - \frac{mv_{o}}{2q\mu_{o}I}\right]$
- $\frac{a}{2}$
- $\mathrm{a}\left\lbrack {1 - \frac{{\mathrm{{mv}}}_{\mathrm{o}}}{{\mathrm{q}}{\mu }_{\mathrm{o}}\mathrm{I}}}\right\rbrack$
- $\mathrm{ae}^{\frac{-4pi m\nu_{o}}{q\mu_{o}I}}$
### Core Logic
Analyzing motion from path phases $\mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{B} \rightarrow \mathrm{C}$ inside the coordinate field:
{{SOL_IMG1}}
$$\vec{\mathrm{B}} = \frac{\mu_0 \mathrm{I}}{2\pi \mathrm{r}}(-\hat{\mathrm{k}})$$
The lorentz magnetic field acceleration rules dictate differential trajectory steps:
$$\int_{v_0}^{0} \frac{v_x dv_x}{\sqrt{v_0^2 - v_x^2}} = -\frac{\mu_0 \mathrm{I q}}{2\pi \mathrm{m}} \int_{a}^{x_1} \frac{dr}{r}$$
Solving this integration step gives the position node parameter:
$$x_1 = a e^{-\frac{2\pi m v_0}{\mu_0 I q}}$$
Compounding this loop interaction for the turning path phase $\mathrm{B} \rightarrow \mathrm{C}$ gives:
### Step 1: Final Solution Integration
$$x = x_1 e^{-\frac{2\pi m v_0}{\mu_0 I q}} = a e^{-\frac{4\pi m v_0}{\mu_0 I q}}$$
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
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### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q22 (2025)
A $4.0\mathrm{~cm}$ long straight wire carrying a current of $8\mathrm{~A}$ is placed perpendicular to an uniform magnetic field of strength $0.15\mathrm{~T}$. The magnetic force on the wire is ________ $\mathrm{mN}$.
### Related Formula
Magnetic force on a current-carrying straight wire:
$$F = I (\vec{L} \times \vec{B}) = I L B \sin \theta$$
where,
$I$ = current,
$L$ = length of the wire,
$B$ = magnetic field strength,
$\theta$ = angle between current direction and magnetic field.
### Core Logic
Given values:
- Length, $L = 4.0\mathrm{~cm} = 0.04\mathrm{~m}$
- Current, $I = 8\mathrm{~A}$
- Magnetic field strength, $B = 0.15\mathrm{~T}$
- Since the wire is placed perpendicular to the field: $\theta = 90^{\circ} \implies \sin 90^{\circ} = 1$
### Step 1: Substitution and Calculation
Substitute values into the force formula:
$$F = 8 \times 0.04 \times 0.15 \times 1$$
$$F = 0.32 \times 0.15 = 0.048\mathrm{~N}$$
Convert this force into millinewtons ($\mathrm{mN}$):
$$F = 0.048 \times 1000 = 48\mathrm{~mN}$$
### Pattern Recognition
Simple, direct application of Bil-sin-theta! Ensure units are converted to standard SI (meters, amperes, tesla) first, and then converted back into millinewtons at the very end to prevent decimal errors.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q24 (2025)
A loop ABCDA, carrying current $I = 12\mathrm{~A}$, is placed in a plane, consists of two semi-circular segments of radius $R_1 = 6\pi\mathrm{~m}$ and $R_2 = 4\pi\mathrm{~m}$. The magnitude of the resultant magnetic field at center O is $k\times 10^{-7}\mathrm{~T}$ The value of k is ________. (Given $\mu_{0}=4\pi\times10^{-7}\mathrm{~T\cdot m\cdot A}^{-1}$)
{{IMG1}}
### Related Formula
Magnetic field at the center of a circular segment of angle $\theta$:
$$B = \frac{\mu_0 I}{4\pi R} \theta$$
For a semicircle ($\theta = \pi$):
$$B_{\text{semi}} = \frac{\mu_0 I}{4 R}$$
### Core Logic
Let's analyze the contributions from each part of the loop to the magnetic field at the center $O$:
- The straight wire segments $AB$ and $CD$ lie along radial lines passing directly through $O$. Since $d\vec{l} \parallel \vec{r}$, their magnetic field contribution is zero:
$$B_{\text{AB}} = B_{\text{CD}} = 0$$
- Semicircular segment of radius $R_2 = 4\pi\mathrm{~m}$ carries current creating a field pointing out of the page (by right-hand rule):
$$B_{R2} = \frac{\mu_0 I}{4 R_2} \quad (\text{Out of page})$$
- Semicircular segment of radius $R_1 = 6\pi\mathrm{~m}$ carries current creating a field pointing into the page:
$$B_{R1} = \frac{\mu_0 I}{4 R_1} \quad (\text{Into page})$$
### Step 1: Calculating Resultant Field
Since $R_2 < R_1$, the field $B_{R2}$ is stronger. The net magnetic field is:
$$B_{\text{net}} = B_{R2} - B_{R1} = \frac{\mu_0 I}{4}\left(\frac{1}{R_2} - \frac{1}{R_1}\right)$$
Substitute the given values ($I = 12\mathrm{~A}$, $R_2 = 4\pi$, $R_1 = 6\pi$, $\mu_0 = 4\pi \times 10^{-7}$):
$$B_{\text{net}} = \frac{(4\pi \times 10^{-7}) \times 12}{4} \left(\frac{1}{4\pi} - \frac{1}{6\pi}\right)$$
$$B_{\text{net}} = 12\pi \times 10^{-7} \times \left(\frac{6\pi - 4\pi}{24\pi^2}\right)$$
$$B_{\text{net}} = 12\pi \times 10^{-7} \times \frac{2\pi}{24\pi^2}$$
$$B_{\text{net}} = 12\pi \times 10^{-7} \times \frac{1}{12\pi} = 1 \times 10^{-7}\mathrm{~T}$$
Comparing this to $k \times 10^{-7}\mathrm{~T}$:
$k = 1$
### Pattern Recognition
Notice how the radial straight wires never contribute to the magnetic field at the center. Semicircles with opposite currents simply subtract. Symmetrizing the math early by factoring out $\frac{\mu_0 I}{4\pi}$ makes the calculations incredibly neat and quick.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Magnetic Effects of Current: Biot-Savart Law
