Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics) — page 4.
Q1149 (2025)
A current of 5A exists in a square loop of side $\frac{1}{\sqrt{2}}\text{ m}$ Then the magnitude of the magnetic field B at the centre of the square loop will be $p\times10^{-6}\text{ T}$ where, value of p is [Take $\mu_{0}=4\pi\times10^{-7}\text{ T mA}^{-1}$].
### Related Formula
The magnetic field $B_{1}$ produced by a straight wire segment carrying current $I$ at a perpendicular distance $d$ is given by the Biot-Savart relation:
$$B_{1} = \frac{\mu_{0}I}{4\pi d}(\sin\theta_{1} + \sin\theta_{2})$$
### Core Logic
As shown in the square geometric layout {{SOL_IMG1}}, the perpendicular distance from any side to the central origin point is exactly half the total side length :
$$d = \frac{a}{2} = \frac{1}{2\sqrt{2}}\text{ m}$$
Connecting the ends of a side to the center forms internal angles of $\theta_{1} = \theta_{2} = 45^{\circ}$.
### Step 1: Summing the Contributions
Calculate the magnetic field contribution from a single side :
$$B_{1} = \frac{10^{-7} \times 5}{\frac{1}{2\sqrt{2}}} \left(\sin 45^{\circ} + \sin 45^{\circ}\right) = 10^{-7} \times 10\sqrt{2} \times \left(\frac{2}{\sqrt{2}}\right) = 2 \times 10^{-6}\text{ T}$$
Since the current flows in the same rotational direction along all four sides, their individual magnetic fields add constructively at the center :
$$B_{\text{net}} = 4 \times B_{1} = 4 \times (2 \times 10^{-6}\text{ T}) = 8 \times 10^{-6}\text{ T}$$
Comparing this with $p \times 10^{-6}\text{ T}$ , we get:
$p = 8$
### Pattern Recognition
The magnetic field at the center of any square loop simplifies to the standard formula: $B = \frac{2\sqrt{2}\mu_{0}I}{\pi a}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q1216 (2025)
An infinite wire has a circular bend of radius $a$ , and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:
{{IMG1}}
- $\frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{\pi}{2} + 1\right]$
- $\frac{\mu_0}{4\pi}\frac{\mathrm{I}}{\mathrm{a}}\left[\frac{3\pi}{2} +1\right]$
- $\frac{\mu_0}{2\pi}\frac{1}{a}\left[\frac{\pi}{2} + 2\right]$
- $\frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{3\pi}{2} + 2\right]$
### Related Formula
The magnetic field contributions from unique structural line elements are given by:
* **Semi-infinite straight wire segment** at a distance perpendicular to its end tip:
$$B_{\text{straight}} = \frac{\mu_0 I}{4\pi a}$$
* **Circular arc path segment** subtending an angle $\theta$ at the center:
$$B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \theta$$
### Core Logic
Let us decompose the structure into three functional parts as mapped out below:
{{SOL_IMG1}}
1. **Segment 1 (Incoming semi-infinite line)**: The straight line extends to infinity, with its terminating tip at a perpendicular distance $a$ from origin $O$. Using the right-hand grip rule, the direction points into the page:
$$B_1 = \frac{\mu_0 I}{4\pi a} \quad (\otimes)$$
2. **Segment 2 (Three-quarter circular loop)**: The loop forms an angle of $\theta = \frac{3\pi}{2}$ radians around $O$. The field points into the page:
$$B_2 = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right) \quad (\otimes)$$
3. **Segment 3 (Outgoing semi-infinite line)**: This line aligns perfectly with the origin $O$ along its vector axis, making $\sin\theta = 0$:
$B_3 = 0$
Summing the total fields via superposition:
$$B = B_1 + B_2 + B_3 = \frac{\mu_0 I}{4\pi a} + \frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2}\right)$$
$$B = \frac{\mu_0 I}{4\pi a} \left[\frac{3\pi}{2} + 1\right]$$
### Pattern Recognition
Always check the axis alignment first. Any straight wire segment whose extended line passes directly through the field point contributes exactly zero to the total magnetic field value.
Q1284 (2025)
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire\'s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
- $\left[\mathrm{a} / 4,3\mathrm{a} / 2\right]$
- $\left[\frac{\mathrm{a}}{2}, 2\mathrm{a}\right]$
- $[\mathrm{a} / 2,3\mathrm{a}]$
- $[\mathrm{a} / 4,2\mathrm{a}]$
### Related Formula
$$B_{\max} = \frac{\mu_0 I}{2\pi a}$$
$$B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2}, \quad B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$$
### Core Logic
The maximum magnetic field occurs right at the wire\'s outer boundary surface ($r=a$) :
$$B_{\max} = \frac{\mu_0 I}{2\pi a}$$
We need positions where $B = \frac{B_{\max}}{2} = \frac{\mu_0 I}{4\pi a}$.
### Step 1: Calculate Inside Distance
$$\frac{\mu_0 I r}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} \implies r = \frac{a}{2}$$
### Step 2: Calculate Outside Distance
$$\frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{4\pi a} \implies r = 2a$$
### Pattern Recognition
Inside the wire, field scales linearly with radius; outside, it falls inversely with radius.
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q1360 (2024)
A galvanometer has a resistance of $50\mathrm{~\Omega}$ and it allows maximum current of $5\mathrm{~mA}$. It can be converted into voltmeter to measure upto $100\mathrm{~V}$ by connecting in series a resistor of resistance:
- $5975\mathrm{~\Omega}$
- $20050\mathrm{~\Omega}$
- $19950\mathrm{~\Omega}$
- $19500\mathrm{~\Omega}$
### Related Formula
Voltmeter series conversion formula:
$$V = I_g(R_g + R)$$
$$R = \frac{V}{I_g} - R_g$$
### Core Logic
Given data:
$R_g = 50\mathrm{~\Omega}$, $I_g = 5\mathrm{~mA} = 5 \times 10^{-3}\mathrm{~A}$, target voltage range $V = 100\mathrm{~V}$.
Substitute values:
$$R = \frac{100}{5 \times 10^{-3}} - 50$$
### Step 1: Complete Arithmetic Evaluation
$R = 20000 - 50 = 19950\mathrm{~\Omega}$
### Pattern Recognition
Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Current Electricity
Q1371 (2024)
A regular polygon of 6 sides is formed by bending a wire of length $4pi$ meter. If an electric current of $4pisqrt{3}\mathrm{~A}$ is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be $x \times 10^{-7}\mathrm{~T}$. The value of $x$ is:
### Related Formula
Magnetic field due to a straight wire segment of length $2L$ at distance $r$:
$$B_1 = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)$$
Total field for a regular hexagon ($n=6$):
$$B = 6 \times B_1$$
### Core Logic
Total perimeter $= 6 \cdot a = 4\pi \implies \text{side length } a = \frac{4\pi}{6} = \frac{2\pi}{3}\mathrm{~m}$.
For a regular hexagon segment, the interior angles relative to the normal vector are $\theta_1 = \theta_2 = 30^\circ$.
The normal distance $r$ from the center to a side is:
$$r = \frac{a}{2} \cot(30^\circ) = \frac{4\pi}{2 \times 6} \times \sqrt{3} = \frac{\sqrt{3}\pi}{3} = \frac{\pi}{\sqrt{3}}\mathrm{~m}$$
### Step 1: Calculate Total Magnetic Field
Substitute $r$ and $I = 4\pi\sqrt{3}\mathrm{~A}$ into the hexagon configuration:
$$B = 6 \times \left[ \frac{\mu_0 I}{4\pi r} (\sin(30^\circ) + \sin(30^\circ)) \right]$$
$$B = 6 \times \left[ \frac{10^{-7} \times 4\pi\sqrt{3}}{\left(\frac{\sqrt{3}\pi}{3}\right)} \times (0.5 + 0.5) \right]$$
$$B = 6 \times \left[ \frac{10^{-7} \times 4\pi\sqrt{3} \times 3}{\sqrt{3}\pi} \times 1 \right]$$
$$B = 6 \times [ 4 \times 3 \times 10^{-7} ] = 6 \times 12 \times 10^{-7} = 72 \times 10^{-7}\mathrm{~T}$$
Thus, $x = 72$.
### Pattern Recognition
For regular polygons, the normal distance $r$ to the side is always $r = \frac{a}{2}\cot(\frac{\pi}{n})$. The contribution from all $n$ symmetric segments adds up constructively.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
