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

Q33 jee_main_2026_21_jan_morning Motion in Magnetic Field

A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :

A. $a = g$ $a = g$
B. $a > g$ $a > g$
C. $a = 0$ $a = 0$
D. $0 < a < g$ $0 < a < g$

Solution

Related Formula

FB = q( v \times B)

Fₙₑₜ = m a

Core Logic

Since the solenoid is placed vertically, the magnetic field $B$ $\vec{B}$ inside the solenoid will be parallel or anti-parallel to the vertical axis (either $+y$ $+y$ or $-y$ $-y$ axis). When the charged particle is released from rest, gravity pulls it vertically downward, meaning it gains velocity $v$ $\vec{v}$ strictly along the $y$ $y$-axis (parallel or anti-parallel to $B$ $\vec{B}$).

Because velocity and magnetic field are collinear ( $v ∥ B$ $\vec{v} \parallel \vec{B}$ or $v ∥ - B$ $\vec{v} \parallel -\vec{B}$), the cross product $v \times B = 0$ $\vec{v} \times \vec{B} = 0$. Therefore, the magnetic force $FB = 0$ $\vec{F}_B = 0$.

Step 1: Calculating Net Acceleration

The only force acting on the particle is gravity.

Fₙₑₜ = m g

$aₙₑₜ = g$ $a_{net} = g$

Motion in Magnetic Field diagram for Q33 - JEE Main 2026 Morning

Pattern Recognition

Whenever a charged particle moves parallel to a magnetic field lines (like moving along the axis of a solenoid), the magnetic force is absolutely zero. It behaves like free fall.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q42 jee_main_2026_21_jan_evening Magnetic Field due to Current Element

An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :

Wire bent into circular loop for Q42 - JEE Main 2026 Evening — Current carrying wire bent into a circular shape of radius r with straight extensions along the x-axis.

A. $(μ₀)/(2π)(I)/(r)(π+1) i$
B. $-(μ₀)/(2π)(I)/(r)(π-1) i$
C. $(μ₀)/(2π)(I)/(r)(π-1) i$
D. $-(μ₀)/(2π)(I)/(r)(π+1) i$

Solution

Related Formula

For a semi-infinite wire segment at distance $r$ $r$:

B = (μ₀ I)/(4π r)

For a full circular loop at its center:

B = (μ₀ I)/(2r)

Core Logic

Vector resolution for Q42 solution - JEE Main 2026 Evening — Current carrying wire bent into a circular shape of radius r with straight extensions along the x-axis.

The total magnetic field at O is the vector sum of fields from three segments:

The incoming semi-infinite wire (AB)
The outgoing semi-infinite wire (DE)
The nearly full circular loop (BCD)

Note: Based on the diagram, the loop is not fully closed, but geometrically it acts as a full circle subtracted by the gap. Typically this standard shape treats the circular part as a full circle and the straight wires as two semi-infinite wires.

BO = BAB + BDE + BBCD

Step 1: Adding the Vector Components

Applying the Right Hand Rule:

Segment AB: current flows along $+x$ $+x$, position vector to O is $+y$ $+y$. $d l \times r = i \times j = k$ . Wait, the diagram shows the loop in the $x-y$ $x-y$ plane. Let's re-examine axes. Based on standard convention, if current is in $xy$ $xy$ plane, field is in $z$ $z$ ( $k$ $\hat{k}$) direction. The solution shows vectors in $i$ $\hat{i}$. This means the axes are drawn such that the loop is in the $y-z$ $y-z$ plane. Yes, the provided axes show $x$ $x$ pointing out, $y$ $y$ to the right, $z$ $z$ upwards.
Segment AB (current along $y$ $y$ axis): $B$ $B$ at origin is along $+x$ $+x$ ( $i$ $\hat{i}$).
Segment DE (current along $y$ $y$ axis): $B$ $B$ at origin is along $+x$ $+x$ ( $i$ $\hat{i}$).
Circular Loop (current clockwise in $y-z$ $y-z$ plane): $B$ $B$ at origin points inwards, i.e., $-x$ $-x$ ( $- i$ $-\hat{i}$).

BAB = (μ₀ I)/(4π r) i

BDE = (μ₀ I)/(4π r) i

BBCD = - (μ₀ I)/(2r) i

Step 2: Final Conclusion

BO = (μ₀ I)/(4π r) i + (μ₀ I)/(4π r) i - (μ₀ I)/(2r) i

BO = (μ₀ I)/(2π r) i - (μ₀ I)/(2r) i

BO = (μ₀ I)/(2π r) (1 - π) i

BO = -(μ₀ I)/(2π r) (π - 1) i

Pattern Recognition

Always separate complex wire geometries into standard segments: infinite wires, semi-infinite wires, and arcs. Use the Right-Hand Rule carefully with the given explicit coordinate frame to avoid sign errors.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q29 jee_main_2026_23_january_evening Magnetic Field due to Current Carrying Wire

The current passing through a conducting loop in the form of equilateral triangle of side

4\sqrt(3) cm

is 2A. The magnetic field at its centroid is

α \times 10⁻⁵T

. The value of

α

is ____. (Given:

μₒ = 4π \times 10⁻⁷

SI units)

A. $2\sqrt(3)$ $2\sqrt{3}$
B. $\sqrt(3)$ $\sqrt{3}$
C. $3\sqrt(3)$ $3\sqrt{3}$
D. $\sqrt(3)2$ $\frac{\sqrt{3}}{2}$

Solution

Related Formula

B = (μ₀ I)/(4π d) [ θ₁ + θ₂]

Core Logic

Magnetic Field due to Current Carrying Wire diagram for Q29 - JEE Main 2026 Evening

For an equilateral triangle, the perpendicular distance $d$ $d$ from centroid to any side is given by $d = a2\sqrt(3)$ $d = \frac{a}{2\sqrt{3}}$, where $a = 4\sqrt(3) cm$ $a = 4\sqrt{3} \, \mathrm{cm}$.

d = 4\sqrt(3)2\sqrt(3) = 2 cm = 2 \times 10⁻² m

The angles subtended by the side at the centroid are $θ₁ = 60°$ $\theta_1 = 60^{\circ}$ and $θ₂ = 60°$ $\theta_2 = 60^{\circ}$.

Step 1: Field due to one side

B₁ = (μ₀)/(4π) \cdot (I)/(d) ( 60° + 60°)

B₁ = 10⁻⁷ \times 22 \times 10⁻² ( \sqrt(3)2 + \sqrt(3)2 )

B₁ = 10⁻⁵ \times \sqrt(3) T

Step 2: Total Magnetic Field

Since there are 3 identical sides and their field vectors point in the same direction at the centroid:

Bₙₑₜ = 3 \times B₁ = 3 \times \sqrt(3) \times 10⁻⁵ T

Comparing with $α \times 10⁻⁵T$ $\alpha \times 10^{-5}\mathrm{T}$, we get $α = 3\sqrt(3)$ $\alpha = 3\sqrt{3}$.

Pattern Recognition

For regular polygons of $n$ $n$ sides, $Bcentroid = n \cdot Bside$ . An equilateral triangle has $n=3$ $n=3$, $d = a/(2\sqrt(3))$ $d = a/(2\sqrt{3})$, and angles are always $60°$ $60^{\circ}$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q26 jee_main_2026_24_january_morning Magnetic Force and Field

Match the List-I with List-II

List-I	List-II
A. Magnetic induction	I. $ML T⁻²A⁻²$ $ML\ T^{-2}A^{-2}$
B. Magnetic flux	II. $ML² T⁻²A⁻²$ $ML^{2}\ T^{-2}A^{-2}$
C. Magnetic permeability	III. $ML⁰ T⁻²A⁻¹$ $ML^{0}\ T^{-2}A^{-1}$
D. Self inductance	IV. $ML² T⁻²A⁻¹$ $ML^{2}\ T^{-2}A^{-1}$

Choose the correct answer from the options given below:

A. $A-IV, B-III, C-I, D-II$ $\text{A-IV, B-III, C-I, D-II}$
B. $A-III, B-IV, C-II, D-I$ $\text{A-III, B-IV, C-II, D-I}$
C. $A-I, B-III, C-IV, D-II$ $\text{A-I, B-III, C-IV, D-II}$
D. $A-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$

Solution

Related Formula

$F = qvB$ $F = qvB$

φ = B \cdot Area

U = (1)/(2) L I²

Core Logic

For Magnetic induction (B):

[B] = [ (F)/(qv) ] = [MT⁻²A⁻¹]

So, A matches III.

For Magnetic Flux ( $φ$ $\phi$):

[φ] = [B] \cdot [Area] = [ML²T⁻²A⁻¹]

So, B matches IV.

For Magnetic Permeability ( $μ$ $\mu$):

[μ] = [MLT⁻²A⁻²]

So, C matches I.

For Self inductance (L): Using $U = (1)/(2) LI²$ $U = \frac{1}{2} LI^{2}$,

[L] = [ML²T⁻²A⁻¹]

Wait, the given option II is $[ML² T⁻² A⁻²]$ $[ML^2 T^{-2} A^{-2}]$. Let's re-verify: Energy $U = [ML² T⁻²]$ $U = [ML^2 T^{-2}]$. $I² = [A²]$ $I^2 = [A^2]$. So $L = [ML² T⁻² A⁻²]$ $L = [ML^2 T^{-2} A^{-2}]$. So, D matches II.

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 $F=qvB$ $F=qvB$ and $U=(1)/(2)LI²$ $U=\frac{1}{2}LI^2$ is the fastest approach.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Electromagnetic Induction

Q32 jee_main_2026_24_january_evening Magnetic Field of Circular Loops

Magnetic Field of Circular Loops diagram for Q32 - JEE Main 2026 Evening — Two identical circular loops positioned parallel to each other with a common central axis O.

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:

A. $3μₒI4\sqrt(2)r toward P$
B. $μₒI4\sqrt(2)r toward P$
C. $μₒI4\sqrt(2)r towards Q$
D. $3μₒI4\sqrt(2)r towards Q$

Solution

Related Formula

B = μ₀ i R²2(x² + R²)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.

Bₙₑₜ = B₁ - B₂

Step 1: Superposition of Fields

Magnetic field due to loop Q (carrying 4I) towards Q, and due to loop P (carrying I) towards P.

Bₙₑₜ = μ₀ (4i) R²2(R² + R²)3/2 - μ₀ (i) R²2(R² + R²)3/2

Bₙₑₜ = 3μ₀ i R²2(2R²)3/2

Step 2: Final Calculation

Bₙₑₜ = 3μ₀ i R²2(2\sqrt(2)R³) = 3μ₀ i4\sqrt(2)R

The direction is towards Q because the field from the loop carrying $4I$ $4I$ 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.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Solution & Explanation

Core Logic

Step 1: Final Solution Integration

Pattern Recognition

Chapter Mix

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