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:

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: This means the angle theta between the velocity vector and the magnetic field vector is either 0 or 180. 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. Magnetic field cannot change the speed of a charged particle, but it changes direction unless v is parallel or anti-parallel to B, in which case the magnetic force vanishes entirely. Class 12 Physics: Moving Charges and Magnetism

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 consta

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: This means the angle theta between the velocity vector and the magnetic field vector is either 0 or 180. 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. Magnetic field cannot change the speed of a cha

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:

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: This means the angle theta between the velocity vector and the magnetic field vector is either 0 or 180. 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. Magnetic field cannot change the speed of a charged particle, but it changes direction unless v is parallel or anti-parallel to B, in which case the magnetic force vanishes entirely. Class 12 Physics: Moving Charges and Magnetism

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 consta

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: This means the angle theta between the velocity vector and the magnetic field vector is either 0 or 180. 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. Magnetic field cannot change the speed of a cha

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion…

Year	2026	2025	2024	Total
Questions	9	18	15	42

Q jee_main_2025_28_jan_evening Biot Savart Law

An infinite wire has a circular bend of radius

a

, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:

Biot Savart Law diagram for Q16 - JEE Main 2025 Evening — An infinite current carrying wire presenting a three-quarter circular bend around center origin O.

A. $(μ₀)/(4π)(I)/(a)[(π)/(2) + 1]$
B. $(μ₀)/(4π) Ia[(3π)/(2) +1]$
C. $(μ₀)/(2π)(1)/(a)[(π)/(2) + 2]$
D. $(μ₀)/(4π)(I)/(a)[(3π)/(2) + 2]$

Solution

Related Formula

The magnetic field contributions from unique structural line elements are given by:

Semi-infinite straight wire segment at a distance perpendicular to its end tip:

Bstraight = (μ₀ I)/(4π a)

Circular arc path segment subtending an angle $θ$ $\theta$ at the center:

Barc = (μ₀ I)/(4π a) θ

Core Logic

Let us decompose the structure into three functional parts as mapped out below:

Biot Savart Law structural analysis diagram for Q16 — An infinite current carrying wire presenting a three-quarter circular bend around center origin O.

Segment 1 (Incoming semi-infinite line): The straight line extends to infinity, with its terminating tip at a perpendicular distance $a$ $a$ from origin $O$ $O$. Using the right-hand grip rule, the direction points into the page:

B₁ = (μ₀ I)/(4π a) ( )

Segment 2 (Three-quarter circular loop): The loop forms an angle of $θ = (3π)/(2)$ $\theta = \frac{3\pi}{2}$ radians around $O$ $O$. The field points into the page:

B₂ = (μ₀ I)/(4π a) ((3π)/(2)) ( )

Segment 3 (Outgoing semi-infinite line): This line aligns perfectly with the origin $O$ $O$ along its vector axis, making $θ = 0$ $\sin\theta = 0$:

$B₃ = 0$ $B_3 = 0$

Summing the total fields via superposition:

B = B₁ + B₂ + B₃ = (μ₀ I)/(4π a) + (μ₀ I)/(4π a)((3π)/(2))

B = (μ₀ I)/(4π a) [(3π)/(2) + 1]

Pattern Recognition

Always check the axis alignment first. Any straight wire segment whose extended line passes directly through the field point contributes exactly zero to the total magnetic field value.

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

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

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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