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

Q56 jee_main_2024_30_january_evening Magnetic Field of a Square Loop

The current of

5 ~A

flows in a square loop of sides

1 ~m

is placed in air. The magnetic field at the centre of the loop is

X\sqrt(2) \times 10⁻⁷ T

. The value of

X

is ________

Numerical Answer. Answer: 40 to 40

Solution

Related Formula

B = (μ₀ i)/(4π d) ( θ₁ + θ₂)

Core Logic

For a square loop of side length $a$ $a$, the perpendicular distance from the center to any side is $d = (a)/(2)$ $d = \frac{a}{2}$. The angles subtended by the ends of one side at the center are $θ₁ = θ₂ = 45^°$ $\theta_1 = \theta_2 = 45^\circ$. Since there are 4 identical sides contributing to the field in the same direction, the total magnetic field is $4$ $4$ times the field due to one side.

Step 1: Calculate Field for One Side

B₁ = (μ₀ i)/(4π (a/2)) ( 45^° + 45^°)

B₁ = 4π \times 10⁻⁷ \times 54π \times 0.5 ( 1\sqrt(2) + 1\sqrt(2) )

B₁ = 10⁻⁷ \times 50.5 \times 2\sqrt(2) = 10 \times 10⁻⁷ \times \sqrt(2)

Step 2: Total Magnetic Field

Btotal = 4 \times B₁ = 4 \times 10\sqrt(2) \times 10⁻⁷ ~T

Btotal = 40\sqrt(2) \times 10⁻⁷ ~T

Comparing with $X\sqrt(2) \times 10⁻⁷ T$ $X\sqrt{2} \times 10^{-7} \mathrm{T}$, we get $X = 40$ $X = 40$.

Pattern Recognition

The magnetic field at the center of an $n$ $n$-sided regular polygon carrying current $I$ $I$ and circumscribed by circle of radius $R$ $R$ is $B = (μ₀ n I)/(2π R) ((π)/(n))$ . Or simply use $B = 4 \times Bwire$ $B = 4 \times B_{\text{wire}}$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q39 jee_main_2024_30_jan_morning Magnetic Field at the Center of a Loop

Two insulated circular loop A and B radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :

Magnetic Field at the Center of a Loop diagram for Q39 - JEE Main 2024 Morning — Two circular current carrying loops placed orthogonally.

A. $\sqrt(2) μ₀ Ia$ $\frac{\sqrt{2} \mu_0 I}{a}$
B. $(μ₀ I)/(2 a)$ $\frac{\mu_0 I}{2 a}$
C. $μ₀ I\sqrt(2) a$ $\frac{\mu_0 I}{\sqrt{2} a}$
D. $(2 μ₀ I)/(a)$ $\frac{2 \mu_0 I}{a}$

Solution

Related Formula

B = (μ₀ I)/(2R)

Bₙₑₜ = \sqrt(B₁² + B₂²) (for mutually perpendicular fields)

Core Logic

The two loops are mutually perpendicular to each other. The magnetic field at the center due to each loop acts along their respective normal axes, meaning the two resulting magnetic field vectors are orthogonal ( $90^°$ $90^\circ$ apart).

Vector resolution showing orthogonal B fields. — Two circular current carrying loops placed orthogonally.

Step 1: Calculate Individual Fields

Magnetic field at the center due to loop A:

BA = (μ₀ I)/(2a)

Magnetic field at the center due to loop B:

BB = (μ₀ I)/(2a)

Step 2: Resultant Magnetic Field

Since $BA$ $B_A$ and $BB$ $B_B$ are perpendicular to each other:

Bₙₑₜ = \sqrt(BA² + BB²)

Bₙₑₜ = \sqrt(((μ₀ I)/(2a))² + ((μ₀ I)/(2a))²)

Bₙₑₜ = (μ₀ I)/(2a) \sqrt(2) = \sqrt(2) μ₀ I2a = μ₀ I\sqrt(2) a

Pattern Recognition

When two identical physical sources generate orthogonal vector fields of magnitude $B$ $B$, the resultant is simply $\sqrt(2)B$ $\sqrt{2}B$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q51 jee_main_2024_30_jan_morning Magnetic Force on Current Carrying Wire

The horizontal component of earth's magnetic field at a place is

3.5 \times 10⁻⁵ ~T

. A very long straight conductor carrying current of

\sqrt(2) ~A

in the direction from South east to North West is placed. The force per unit length experienced by the conductor is

\times 10⁻⁶ ~N/m

.

Numerical Answer. Answer: 35 to 35

Solution

Related Formula

F = I L B θ

(F)/(L) = I B θ

Core Logic

Earth's horizontal magnetic field ( $BH$ $B_H$) points strictly from South to North. The conductor is aligned from South-East to North-West. Thus, the angle $θ$ $\theta$ between the current flow direction (SE to NW) and Earth's magnetic field (S to N) is exactly $45^°$ $45^\circ$.

Step 1: Compute Force per Unit Length

$BH = 3.5 \times 10⁻⁵ ~T$ $B_H = 3.5 \times 10^{-5} \mathrm{~T}$ $i = \sqrt(2) ~A$ $i = \sqrt{2} \mathrm{~A}$ $θ = 45^°$ $\theta = 45^\circ$

(F)/( ) = i BH θ

(F)/( ) = \sqrt(2) \times (3.5 \times 10⁻⁵) \times 1\sqrt(2)

(F)/( ) = 3.5 \times 10⁻⁵ ~N/m

(F)/( ) = 35 \times 10⁻⁶ ~N/m

Step 2: Extract Value

The required multiplier for $10⁻⁶$ $10^{-6}$ is 35.

Pattern Recognition

Geographical alignments often yield clean angles like $45^°$ $45^\circ$ or $90^°$ $90^\circ$. S-N field intersecting SE-NW current creates a $45^°$ $45^\circ$ angular intersection, perfectly canceling the $\sqrt(2)$ $\sqrt{2}$ amp current.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Magnetism and Matter

Q32 jee_main_2024_31_jan_evening Torque on a Current Loop

A uniform magnetic field of

2 \times 10⁻³ T

acts along positive Y-direction. A rectangular loop of sides

20 cm

and

10 cm

with current of

5 A

is Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :

A. $2 \times 10⁻⁴ N-m$ $2 \times 10^{-4} \text{ N-m}$ along positive Z-direction
B. $2 \times 10⁻⁴ N-m$ $2 \times 10^{-4} \text{ N-m}$ along negative Z-direction
C. $2 \times 10⁻⁴ N-m$ $2 \times 10^{-4} \text{ N-m}$ along positive X-direction
D. $2 \times 10⁻⁴ N-m$ $2 \times 10^{-4} \text{ N-m}$ along positive Y-direction

Solution

Related Formula

M = i A

τ = M \times B

Core Logic

The area vector $A$ $\vec{A}$ is perpendicular to the plane of the loop. According to the right-hand rule, an anticlockwise current observed from the negative X-axis means the area vector points in the negative X-direction.

Torque on a Current Loop diagram for Q32 - JEE Main 2024 Evening

Step 1: Calculating Magnetic Moment

Area of the loop, $A = 0.2 m \times 0.1 m = 0.02 m²$

Magnetic moment vector:

M = i A

M = 5 \times (0.2) \times (0.1) (- i)

M = 0.1(- i) A m²

Step 2: Calculating Torque

The magnetic field is $B = 2 \times 10⁻³ j T$ .

Torque on the loop:

τ = M \times B

τ = 0.1(- i) \times (2 \times 10⁻³ j)

τ = 2 \times 10⁻⁴ (- k) N-m

Step 3: Final Direction

The $- k$ $-\hat{k}$ direction corresponds to the negative Z-direction. Magnitude is $2 \times 10⁻⁴ N-m$ $2 \times 10^{-4} \text{ N-m}$.

Pattern Recognition

Identify the plane of the loop to get the possible normal vectors. Since the loop is in the Y-Z plane, $A$ $\vec{A}$ must be $\pm i$ $\pm\hat{i}$. 'Anticlockwise from -X' fixes it as $- i$ $-\hat{i}$. Simply apply cross product $- i \times j = - k$ $-\hat{i} \times \hat{j} = -\hat{k}$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q56 jee_main_2024_31_jan_evening Magnetic Field at the Center of a Coil

Two circular coils P and Q of 100 turns each have same radius of

π cm

. The currents in P and Q are 1 A and 2 A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is

\sqrt(x) mT

, where

x =

[Use

μ₀ = 4π \times 10⁻⁷ T m A⁻¹

]

Numerical Answer. Answer: 20 to 20

Solution

Related Formula

B = (μ₀ N i)/(2r)

$Bₙₑₜ = \sqrt(B₁² + B₂²)$ (for mutually perpendicular planes)

Core Logic

Calculate the magnetic field generated by each coil at their common center. Since the planes of the coils are perpendicular, their axial magnetic field vectors are also mutually perpendicular.

Magnetic Field at the Center of a Coil diagram for Q56 - JEE Main 2024 Evening

Step 1: Calculate Individual Fields

Radius $r = π cm = π \times 10⁻² m$ . $N = 100$ $N = 100$ $iP = 1 A$ $i_P = 1 \text{ A}$, $iQ = 2 A$ $i_Q = 2 \text{ A}$.

For coil P:

BP = (μ₀ N iP)/(2r) = (4π \times 10⁻⁷) \times 100 \times 12 \times π \times 10⁻²

BP = 4 \times 10⁻⁵2 \times 10⁻² = 2 \times 10⁻³ T = 2 mT

For coil Q:

BQ = (μ₀ N iQ)/(2r) = (4π \times 10⁻⁷) \times 100 \times 22 \times π \times 10⁻² = 4 \times 10⁻³ T = 4 mT

Step 2: Calculate Resultant Field

Bₙₑₜ = \sqrt(BP² + BQ²)

Bₙₑₜ = \sqrt(2² + 4²) = \sqrt(4 + 16) = \sqrt(20) mT

Step 3: Extract x

Given format is $\sqrt(x) mT$ $\sqrt{x} \text{ mT}$, so $x = 20$ $x = 20$.

Pattern Recognition

For concentric perpendicular coils of same radius and turns, $Bₙₑₜ = B₀ \sqrt(i₁² + i₂²)$ where $B₀$ $B_0$ is the base field for 1 A. This speeds up calculation considerably.

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 7

Solution

Related Formula

Core Logic

Step 1: Calculate Field for One Side

Step 2: Total Magnetic Field

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Individual Fields

Step 2: Resultant Magnetic Field

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Force per Unit Length

Step 2: Extract Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Magnetic Moment

Step 2: Calculating Torque

Step 3: Final Direction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Individual Fields

Step 2: Calculate Resultant Field

Step 3: Extract x

Pattern Recognition

Chapter Mix

More Moving Charges and Magnetism Questions — jee_main_2025_28_jan_morning

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