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 v0 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 x from the wire. The value of x is [mu0 is vacuum permeability]

Analyzing motion from path phases A arrow B and B arrow C inside the coordinate field: The lorentz magnetic field acceleration rules dictate differential trajectory steps: Solving this integration step gives the position node parameter: Compounding this loop interaction for the turning path phase B arrow C gives: Matches criteria for option (4). Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work magnetic operations. Class 12 Physics: Moving Charges and Magnetism

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 v0 along the direction of current

Analyzing motion from path phases A arrow B and B arrow C inside the coordinate field: The lorentz magnetic field acceleration rules dictate differential trajectory steps: Solving this integration step gives the position node parameter: Compounding this loop interaction for the turning path phase B arrow C gives: Matches criteria for option (4). Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work mag

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 v0 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 x from the wire. The value of x is [mu0 is vacuum permeability]

Analyzing motion from path phases A arrow B and B arrow C inside the coordinate field: The lorentz magnetic field acceleration rules dictate differential trajectory steps: Solving this integration step gives the position node parameter: Compounding this loop interaction for the turning path phase B arrow C gives: Matches criteria for option (4). Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work magnetic operations. Class 12 Physics: Moving Charges and Magnetism

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 v0 along the direction of current

Analyzing motion from path phases A arrow B and B arrow C inside the coordinate field: The lorentz magnetic field acceleration rules dictate differential trajectory steps: Solving this integration step gives the position node parameter: Compounding this loop interaction for the turning path phase B arrow C gives: Matches criteria for option (4). Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work mag

Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge "q" is…

Year	2026	2025	2024	Total
Questions	9	13	15	37

Q jee_main_2025_24_jan_morning Magnetic Field due to a Current Element

A current of 5A exists in a square loop of side

1\sqrt(2) m

Then the magnitude of the magnetic field B at the centre of the square loop will be

p\times10⁻⁶ T

where, value of p is [Take

μ₀=4π\times10⁻⁷ T mA⁻¹

].

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

The magnetic field $B₁$ $B_{1}$ produced by a straight wire segment carrying current $I$ $I$ at a perpendicular distance $d$ $d$ is given by the Biot-Savart relation:

B₁ = μ₀I4π d( θ₁ + θ₂)

Core Logic

As shown in the square geometric layout

Magnetic Field due to a Current Element diagram for Q24 - JEE Main 2025 Morning

, the perpendicular distance from any side to the central origin point is exactly half the total side length :

d = (a)/(2) = 12\sqrt(2) m

Connecting the ends of a side to the center forms internal angles of $θ₁ = θ₂ = 45°$ $\theta_{1} = \theta_{2} = 45^{\circ}$.

Step 1: Summing the Contributions

Calculate the magnetic field contribution from a single side :

B₁ = 10⁻⁷ \times 5 12\sqrt(2) ( 45° + 45°) = 10⁻⁷ \times 10\sqrt(2) \times ( 2\sqrt(2)) = 2 \times 10⁻⁶ T

Since the current flows in the same rotational direction along all four sides, their individual magnetic fields add constructively at the center :

Bₙₑₜ = 4 \times B₁ = 4 \times (2 \times 10⁻⁶ T) = 8 \times 10⁻⁶ T

Comparing this with $p \times 10⁻⁶ T$ $p \times 10^{-6}\text{ T}$ , we get:

$p = 8$ $p = 8$

Pattern Recognition

The magnetic field at the center of any square loop simplifies to the standard formula: $B = 2\sqrt(2)μ₀Iπ a$ $B = \frac{2\sqrt{2}\mu_{0}I}{\pi a}$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q jee_main_2025_29_jan_morning Ampere\'s Circuital Law

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

A. $[a / 4,3a / 2]$
B. $[ a2, 2a]$
C. $[a / 2,3a]$ $[\mathrm{a} / 2,3\mathrm{a}]$
D. $[a / 4,2a]$ $[\mathrm{a} / 4,2\mathrm{a}]$

Solution

Related Formula

B = (μ₀ I)/(2π a)

Bᵢₙ = (μ₀ I r)/(2π a²), Bout = (μ₀ I)/(2π r)

Core Logic

The maximum magnetic field occurs right at the wire\'s outer boundary surface ( $r=a$ $r=a$) :

B = (μ₀ I)/(2π a)

We need positions where $B = B2 = (μ₀ I)/(4π a)$ .

Step 1: Calculate Inside Distance

(μ₀ I r)/(2π a²) = (μ₀ I)/(4π a) r = (a)/(2)

Step 2: Calculate Outside Distance

(μ₀ I)/(2π r) = (μ₀ I)/(4π a) 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

Q jee_main_2024_01_february_morning Galvanometer Conversion

A galvanometer has a resistance of

50~Ω

and it allows maximum current of

5~mA

. It can be converted into voltmeter to measure upto

100~V

by connecting in series a resistor of resistance:

A. $5975~Ω$ $5975\mathrm{~\Omega}$
B. $20050~Ω$ $20050\mathrm{~\Omega}$
C. $19950~Ω$ $19950\mathrm{~\Omega}$
D. $19500~Ω$ $19500\mathrm{~\Omega}$

Solution

Related Formula

Voltmeter series conversion formula:

V = Ig(Rg + R)

R = (V)/(Ig) - Rg

Core Logic

Given data: $Rg = 50~Ω$ $R_g = 50\mathrm{~\Omega}$, $Ig = 5~mA = 5 \times 10⁻³~A$ , target voltage range $V = 100~V$ $V = 100\mathrm{~V}$.

Substitute values:

R = 1005 \times 10⁻³ - 50

Step 1: Complete Arithmetic Evaluation

$R = 20000 - 50 = 19950~Ω$

Pattern Recognition

Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Current Electricity

Q jee_main_2024_01_february_morning Magnetic Field due to a Current Element

A regular polygon of 6 sides is formed by bending a wire of length

4pi

meter. If an electric current of

4pisqrt3~A

is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be

x \times 10⁻⁷~T

. The value of

x

is:

Numerical Answer. Answer: 72 to 72

Solution

Related Formula

Magnetic field due to a straight wire segment of length $2L$ $2L$ at distance $r$ $r$:

B₁ = (μ₀ I)/(4π r)( θ₁ + θ₂)

Total field for a regular hexagon ( $n=6$ $n=6$):

B = 6 \times B₁

Core Logic

Total perimeter $= 6 \cdot a = 4π side length a = (4π)/(6) = (2π)/(3)~m$ . For a regular hexagon segment, the interior angles relative to the normal vector are $θ₁ = θ₂ = 30^°$ $\theta_1 = \theta_2 = 30^\circ$.

The normal distance $r$ $r$ from the center to a side is:

r = (a)/(2) (30^°) = (4π)/(2 \times 6) \times \sqrt(3) = \sqrt(3)π3 = π\sqrt(3)~m

Step 1: Calculate Total Magnetic Field

Substitute $r$ $r$ and $I = 4π\sqrt(3)~A$ $I = 4\pi\sqrt{3}\mathrm{~A}$ into the hexagon configuration:

B = 6 \times [ (μ₀ I)/(4π r) ( (30^°) + (30^°)) ]

B = 6 \times [ 10⁻⁷ \times 4π\sqrt(3)( \sqrt(3)π3) \times (0.5 + 0.5) ]

B = 6 \times [ 10⁻⁷ \times 4π\sqrt(3) \times 3\sqrt(3)π \times 1 ]

B = 6 \times [ 4 \times 3 \times 10⁻⁷ ] = 6 \times 12 \times 10⁻⁷ = 72 \times 10⁻⁷~T

Thus, $x = 72$ $x = 72$.

Pattern Recognition

For regular polygons, the normal distance $r$ $r$ to the side is always $r = (a)/(2) ((π)/(n))$ $r = \frac{a}{2}\cot(\frac{\pi}{n})$. The contribution from all $n$ $n$ symmetric segments adds up constructively.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q39 jee_main_2024_29_january_evening Motion of Charged Particle in Magnetic Field

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii

R₁

and

R₂

respectively. The mass ratio of X and Y is:

A. $((R₂)/(R₁))²$ $\left(\frac{R_2}{R_1}\right)^2$
B. $((R₁)/(R₂))²$ $\left(\frac{R_1}{R_2}\right)^2$
C. $((R₁)/(R₂))$ $\left(\frac{R_1}{R_2}\right)$
D. $((R₂)/(R₁))$ $\left(\frac{R_2}{R_1}\right)$

Solution

Related Formula

The radius $R$ $R$ of the path of a charged particle moving perpendicular to a magnetic field $B$ $B$ is:

R = (mv)/(qB) = (p)/(qB)

In terms of kinetic energy $K$ $K$:

R = \sqrt(2mK)qB

Since the particle is accelerated through potential $V$ $V$, kinetic energy $K = qV$ $K = qV$:

R = \sqrt(2mqV)qB

R = (1)/(B) \sqrt((2mV)/(q))

Core Logic

For both particles X and Y, the following parameters are the same:

Potential Difference, $V$ $V$
Magnetic Field, $B$ $B$
Charge, $q$ $q$

Therefore, we have the proportionality:

R \propto \sqrt(m) R² \propto m

Step 1: Calculate Mass Ratio

Using the proportionality relationship:

(m₁)/(m₂) = ( (R₁)/(R₂) )²

Thus, the mass ratio of X and Y is $((R₁)/(R₂))²$ $\left(\frac{R_1}{R_2}\right)^2$.

Pattern Recognition

Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as $R \propto \sqrt(m)$ $R \propto \sqrt{m}$. Squaring both sides yields $m \propto R²$ $m \propto R^2$ instantly.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Solution & Explanation

Core Logic

Step 1: Final Solution Integration

Pattern Recognition

Chapter Mix

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Summing the Contributions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Inside Distance

Step 2: Calculate Outside Distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Complete Arithmetic Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Total Magnetic Field

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Mass Ratio

Pattern Recognition

Chapter Mix

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