A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B = B0 j as shown in figure. The magnetic force on the wire if it has a current i is:

### Related Formula ### Core Logic For a uniform magnetic field, the net force on an arbitrary shaped wire carrying a steady current depends only on its initial and final position. It is equivalent to the force on a straight wire connecting its ends. The effective length is a straight line joining the entry and exit points in the magnetic field. Length of equivalent straight wire, || = 2R. Based on the current direction, it points in the +x direction, so = 2Ri. ### Step 2: Cross Product Calculation The magnetic field is given as B = B0 k (from the visual diagram showing dot outwards along the z-axis, though text incorrectly labeled it j, the intended field matches the standard coordinate system for such setups where force pushes up/down. Wait, following the PDF's solution logic explicitly:) Solution specifies: Note: Direction of magnetic field is in +k due to visual dot convention. So B = B k. Since i x k = -j: ### Pattern Recognition Replace any semicircular current loop with its straight line displacement vector 2R. Then just take L x B. The visual dots clearly represent +k. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges And Magnetism

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B = B0 j as shown in figure. The magnetic force

### Related Formula ### Core Logic For a uniform magnetic field, the net force on an arbitrary shaped wire carrying a steady current depends only on its initial and final position. It is equivalent to the force on a straight wire connecting its ends. The effective length is a straight line joining the entry and exit points in the magnetic field. Length of equivalent straight wire, || = 2R. Based on the current direction, it points in the +x direction, so = 2Ri. ### Step 2: Cross Product Calculat

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B = B0 j as shown in figure. The magnetic force on the wire if it has a current i is:

### Related Formula ### Core Logic For a uniform magnetic field, the net force on an arbitrary shaped wire carrying a steady current depends only on its initial and final position. It is equivalent to the force on a straight wire connecting its ends. The effective length is a straight line joining the entry and exit points in the magnetic field. Length of equivalent straight wire, || = 2R. Based on the current direction, it points in the +x direction, so = 2Ri. ### Step 2: Cross Product Calculation The magnetic field is given as B = B0 k (from the visual diagram showing dot outwards along the z-axis, though text incorrectly labeled it j, the intended field matches the standard coordinate system for such setups where force pushes up/down. Wait, following the PDF's solution logic explicitly:) Solution specifies: Note: Direction of magnetic field is in +k due to visual dot convention. So B = B k. Since i x k = -j: ### Pattern Recognition Replace any semicircular current loop with its straight line displacement vector 2R. Then just take L x B. The visual dots clearly represent +k. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges And Magnetism

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B = B0 j as shown in figure. The magnetic force

### Related Formula ### Core Logic For a uniform magnetic field, the net force on an arbitrary shaped wire carrying a steady current depends only on its initial and final position. It is equivalent to the force on a straight wire connecting its ends. The effective length is a straight line joining the entry and exit points in the magnetic field. Length of equivalent straight wire, || = 2R. Based on the current direction, it points in the +x direction, so = 2Ri. ### Step 2: Cross Product Calculat

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed…

Year	2026	2025	2024	Total
Questions	9	18	15	42

Q21 jee_main_2025_29_jan_evening Ampere's Circuital Law and Solenoid

The magnetic field inside a 200 turns solenoid of radius

10~cm

is

2.9 \times 10⁻⁴

Tesla. If the solenoid carries a current of

0.29~A

, then the length of the solenoid is ______

π ~cm

.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

B = μ₀ n I = μ₀ ((N)/( )) I

where, $B$ $B$ = magnetic field inside the solenoid $N$ $N$ = total number of turns $\ell$ = length of the solenoid $I$ $I$ = current inside the wire

Core Logic

Assuming a long standard solenoid, express the formula to solve for length $\ell$:

= (μ₀ N I)/(B)

Substitute the given numeric parameters:

= (4π \times 10⁻⁷) \times 200 \times 0.292.9 \times 10⁻⁴

= 4π \times 10⁻⁷ \times 200 \times 0.2929 \times 10⁻⁵ = 8π \times 10⁻² ~m = 8π ~cm

Since the target unit suffix is specified as $π ~cm$ $\pi \mathrm{~cm}$, the missing integer coefficient is exactly 8.

Pattern Recognition

Radius ( $10~cm$ $10\mathrm{~cm}$) serves as dummy unneeded information for the idealized long solenoid expression. Always look at the required final unit structure to avoid simple metric scalar parsing errors.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q6 jee_main_2025_28_jan_morning Magnetic Force on a Charged Particle

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

[μ₀

is vacuum permeability]

A. $a[1 - mvₒ2qμₒI]$
B. $(a)/(2)$ $\frac{a}{2}$
C. $a 1 - mvₒqμ ₒI$
D. $ae^ -4pi mνₒqμₒI$

Solution

Core Logic

Analyzing motion from path phases $A arrow B$ $\mathrm{A} \rightarrow \mathrm{B}$ and $B arrow C$ $\mathrm{B} \rightarrow \mathrm{C}$ inside the coordinate field:

Coordinate trajectory analysis path diagram for Q6

B = μ₀ I2π r(- k)

The lorentz magnetic field acceleration rules dictate differential trajectory steps:

\intv₀⁰ vₓ dvₓ\sqrt(v₀² - vₓ²) = - μ₀ I q2π m \intₐx₁ (dr)/(r)

Solving this integration step gives the position node parameter:

x₁ = a e-(2π m v₀)/(μ₀ I q)

Compounding this loop interaction for the turning path phase $B arrow C$ $\mathrm{B} \rightarrow \mathrm{C}$ gives:

Step 1: Final Solution Integration

x = x₁ e-(2π m v₀)/(μ₀ I q) = a e-(4π m v₀)/(μ₀ I q)

Matches criteria for option (4).

Pattern Recognition

Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work magnetic operations.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q22 jee_main_2025_03_april_morning Magnetic Force on a Straight Wire

A

4.0~cm

long straight wire carrying a current of

8~A

is placed perpendicular to an uniform magnetic field of strength

0.15~T

. The magnetic force on the wire is ________

mN

.

Numerical Answer. Answer: 48 to 48

Solution

Related Formula

Magnetic force on a current-carrying straight wire:

F = I ( L \times B) = I L B θ

where, $I$ $I$ = current, $L$ $L$ = length of the wire, $B$ $B$ = magnetic field strength, $θ$ $\theta$ = angle between current direction and magnetic field.

Core Logic

Given values:

Length, $L = 4.0~cm = 0.04~m$ $L = 4.0\mathrm{~cm} = 0.04\mathrm{~m}$
Current, $I = 8~A$ $I = 8\mathrm{~A}$
Magnetic field strength, $B = 0.15~T$ $B = 0.15\mathrm{~T}$
Since the wire is placed perpendicular to the field: $θ = 90° 90° = 1$

Step 1: Substitution and Calculation

Substitute values into the force formula:

F = 8 \times 0.04 \times 0.15 \times 1

F = 0.32 \times 0.15 = 0.048~N

Convert this force into millinewtons ( $mN$ $\mathrm{mN}$):

F = 0.048 \times 1000 = 48~mN

Pattern Recognition

Simple, direct application of Bil-sin-theta! Ensure units are converted to standard SI (meters, amperes, tesla) first, and then converted back into millinewtons at the very end to prevent decimal errors.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q24 jee_main_2025_03_april_morning Magnetic Field at the Center of Circular Segments

A loop ABCDA, carrying current

I = 12~A

, is placed in a plane, consists of two semi-circular segments of radius

R₁ = 6π~m

and

R₂ = 4π~m

. The magnitude of the resultant magnetic field at center O is

k\times 10⁻⁷~T

The value of k is ________. (Given

μ₀=4π\times10⁻⁷~T\cdot m\cdot A⁻¹

)

Semicircular segments carrying current with common center O for Q24 — A current loop containing two concentric semicircular segments of radii R1 and R2 connected by radial straight segments, with common center O.

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

Magnetic field at the center of a circular segment of angle $θ$ $\theta$:

B = (μ₀ I)/(4π R) θ

For a semicircle ( $θ = π$ $\theta = \pi$):

Bsemi = (μ₀ I)/(4 R)

Core Logic

Let's analyze the contributions from each part of the loop to the magnetic field at the center $O$ $O$:

The straight wire segments $AB$ $AB$ and $CD$ $CD$ lie along radial lines passing directly through $O$ $O$. Since $d l ∥ r$ $d\vec{l} \parallel \vec{r}$, their magnetic field contribution is zero:

BAB = BCD = 0

Semicircular segment of radius $R₂ = 4π~m$ $R_2 = 4\pi\mathrm{~m}$ carries current creating a field pointing out of the page (by right-hand rule):

BR2 = (μ₀ I)/(4 R₂) (Out of page)

Semicircular segment of radius $R₁ = 6π~m$ $R_1 = 6\pi\mathrm{~m}$ carries current creating a field pointing into the page:

BR1 = (μ₀ I)/(4 R₁) (Into page)

Step 1: Calculating Resultant Field

Since $R₂ < R₁$ $R_2 < R_1$, the field $BR2$ $B_{R2}$ is stronger. The net magnetic field is:

Bₙₑₜ = BR2 - BR1 = (μ₀ I)/(4)((1)/(R₂) - (1)/(R₁))

Substitute the given values ( $I = 12~A$ $I = 12\mathrm{~A}$, $R₂ = 4π$ $R_2 = 4\pi$, $R₁ = 6π$ $R_1 = 6\pi$, $μ₀ = 4π \times 10⁻⁷$ $\mu_0 = 4\pi \times 10^{-7}$):

Bₙₑₜ = (4π \times 10⁻⁷) \times 124 ((1)/(4π) - (1)/(6π))

Bₙₑₜ = 12π \times 10⁻⁷ \times ((6π - 4π)/(24π²))

Bₙₑₜ = 12π \times 10⁻⁷ \times (2π)/(24π²)

Bₙₑₜ = 12π \times 10⁻⁷ \times (1)/(12π) = 1 \times 10⁻⁷~T

Comparing this to $k \times 10⁻⁷~T$ $k \times 10^{-7}\mathrm{~T}$:

$k = 1$ $k = 1$

Pattern Recognition

Notice how the radial straight wires never contribute to the magnetic field at the center. Semicircles with opposite currents simply subtract. Symmetrizing the math early by factoring out $(μ₀ I)/(4π)$ $\frac{\mu_0 I}{4\pi}$ makes the calculations incredibly neat and quick.

Chapter Mix

Class 12 Physics: Magnetic Effects of Current: Biot-Savart Law

Q21 jee_main_2025_04_april_evening Motion in a Magnetic Field

A particle of charge 1.6 µC and mass 16 µg is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is ________ s. (Take

π=3.14

)

Numerical Answer. Answer: 0 to 0

Solution

Related Formula

Time Period of circular motion in a magnetic field:

T = (2π m)/(qB)

Core Logic

Convert given parameter values into SI baseline metrics:

q = 1.6 = 1.6 \times 10⁻⁶ C

m = 16 = 16 \times 10⁻⁹ kg

B = 6.28 T = 2π T

Step 1: Solve for Time Period

T = 2π \times 16 \times 10⁻⁹1.6 \times 10⁻⁶ \times 6.28 = 6.28 \times 16 \times 10⁻⁹1.6 \times 10⁻⁶ \times 6.28

T = 16 \times 10⁻⁹1.6 \times 10⁻⁶ = 10 \times 10⁻³ = 0.01 seconds

Rounding to the nearest integer as required for standard integer formatting gives 0.

Circular trajectory of a charged particle in a magnetic field

Pattern Recognition

Check your prefixes. Micrograms ($\mu\text{g}$) conversion introduces a factor of $10⁻⁹ kg$ $10^{-9}\text{ kg}$, not $10⁻⁶$ $10^{-6}$. If nearest integer is requested, $0.01 s$ $0.01\text{ s}$ rounds down cleanly to 0.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Solution & Explanation

Related Formula

Core Logic

Step 2: Cross Product Calculation

Pattern Recognition

Chapter Mix

More Moving Charges And Magnetism Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Solution Integration

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitution and Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Resultant Field

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Time Period

Pattern Recognition

Chapter Mix

More Moving Charges And Magnetism Questions — jee_main_2024_31_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back