A loop ABCDA, carrying current I = 12~A, is placed in a plane, consists of two semi-circular segments of radius R1 =…

Q3 jee_main_2025_24_jan_evening Magnetic Force on a Charge

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path. Reason (R): The magnetic field in that region is along the direction of velocity of the electron. In the light of the above statements, choose the correct answer from the options given below:

A. $(A)$ $(A)$ is false but $(R)$ $(R)$ is true
B. Both $(A)$ $(A)$ and $(R)$ $(R)$ are true and $(R)$ $(R)$ is the correct explanation of $(A)$ $(A)$
C. Both $(A)$ $(A)$ and $(R)$ $(R)$ are true but $(R)$ $(R)$ is NOT the correct explanation of $(A)$ $(A)$
D. $(A)$ $(A)$ is true but $(R)$ $(R)$ is false

Solution

Related Formula

F = q( v \times B)

Core Logic

For a particle to move with a constant velocity in a straight line inside a magnetic field alone, the net magnetic force must be zero:

F = 0 v ∥ B

This means the angle $θ$ $\theta$ between the velocity vector and the magnetic field vector is either $0°$ $0^{\circ}$ or $180°$ $180^{\circ}$. Thus, if the magnetic field is along the direction of velocity, the force is zero, allowing unaccelerated straight-line motion. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Pattern Recognition

Magnetic field cannot change the speed of a charged particle, but it changes direction unless $v$ $\vec{v}$ is parallel or anti-parallel to $B$ $\vec{B}$, in which case the magnetic force vanishes entirely.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q5 jee_main_2025_24_jan_evening Ampere's Circuital Law

A long straight wire of a circular cross-section with radius 'a' carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by:

A. Graph (1)
B. Graph (2)
C. Graph (3)
D. Graph (4)

Solution

Related Formula

Bᵢₙ = (μ₀ I r)/(2π a²) Bᵢₙ \propto r

Bout = (μ₀ I)/(2π r) Bout \propto (1)/(r)

Core Logic

Inside the wire ( $r < a$ $r < a$), the magnetic field grows linearly with distance $r$ $r$ from the axis. At the surface ( $r = a$ $r = a$), it reaches its maximum value $B = (μ₀ I)/(2π a)$ $B_{\max} = \frac{\mu_0 I}{2\pi a}$. Outside the wire ( $r > a$ $r > a$), it decays inversely with $r$ $r$. This combination matches the curve shown in Graph (1).

Ampere law plot variation for thick wire Q5 — magnetic field variation, ampere law plot, current carrying wire

Pattern Recognition

Solid cylinder current profile: linear inside ( $B \propto r$ $B \propto r$), hyperbolic outside ( $B \propto 1/r$ $B \propto 1/r$).

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q11 jee_main_2025_24_jan_evening Ampere's Circuital Law

N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity

ω

as shown in the figure

Rotating charge ring with Amperian loops Q11 — The figure illustrates a rotating ring of charges with two distinct Amperian paths A and B intersecting the ring path.

. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents,

IA - IB

, for the given Amperian loops is

A. $N²2πqω$ $\frac{N^{2}}{2\pi}q\omega$
B. $(2π)/(N)qω$ $\frac{2\pi}{N}q\omega$
C. $(N)/(2π)qω$ $\frac{N}{2\pi}q\omega$
D. $(N)/(π)qω$ $\frac{N}{\pi}q\omega$

Solution

Related Formula

I = (q)/(T) = (qω)/(2π)

Core Logic

The loop A encloses one of the moving point charges as it moves past, giving a current contribution localized to that cross-sectional segment intersection:

IA = (Nq)/(((2π)/(ω))) = (Nqω)/(2π)

Loop B encloses the entire loop surface coplanar or enclosing the ring structure fully without clipping individual passing current tracks perpendicularly in the same directional fashion, resulting in zero net cross-surface passing enclosed current:

$IB = 0$ $I_B = 0$

Therefore, the difference is: $IA - IB = (Nqω)/(2π)$ $$I_A - I_B = \frac{Nq\omega}{2\pi}$$

Enclosed current lines interpretation schematic Q11 — The figure illustrates a rotating ring of charges with two distinct Amperian paths A and B intersecting the ring path.

Pattern Recognition

A current loop has net passing current across a large overarching bounding box equal to zero if it doesn't cross the boundary surfaces symmetrically.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q21 jee_main_2025_24_jan_evening Solenoid

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75ns. The number of turns per metre in the solenoid is ____.

Solenoid cross section with internal electron circular orbit Q21 — The figure details a thick solenoid cylinder with a internal cross section displaying a charge tracking loop.

[Take mass of electron

mₑ = 9 \times 10⁻³¹

kg, charge of electron

|qₑ| = 1.6 \times 10⁻¹⁹

C,

μ₀ = 4π \times 10⁻⁷ (N)/(A²), 1 ns = 10⁻⁹ s

]

Numerical Answer. Answer: 250 to 250

Solution

Related Formula

Time period of a revolving charge in a magnetic field:

T = (2π m)/(qB)

Magnetic field inside a long solenoid: $B = μ₀ n I$ $B = \mu_0 n I$

Core Logic

Combining the expressions to isolate $n$ $n$ (turns per meter):

T = (2π m)/(q(μ₀ n I))

Substituting the given constants:

75 \times 10⁻⁹ = 2π \times 9 \times 10⁻³¹1.6 \times 10⁻¹⁹ \times 4π \times 10⁻⁷ \times n \times 1.5

Simplifying terms:

75 \times 10⁻⁹ = 18π \times 10⁻³¹9.6π \times 10⁻²⁶ \times n = 1.875 \times 10⁻⁵n

n = 1.875 \times 10⁻⁵75 \times 10⁻⁹ = 250

Pattern Recognition

The circular motion time period depends exclusively on the field magnitude $B$ $B$, completely independent of the orbit's velocity or radius.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculating Resultant Field

Pattern Recognition

Chapter Mix

Reference Study Guides

More Magnetic Effects of Current Previous-Year Questions — Page 5

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Related Formula

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Pattern Recognition

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Related Formula

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Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Summing the Contributions

Pattern Recognition

Chapter Mix

More Magnetic Effects of Current Questions — jee_main_2025_03_april_morning

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