Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If oxygen ion (O-2) and Hydrogen ion (H+) enter normal to the magnetic field with equal momentum, then the path of O-2 ion has a smaller curvature than that of H. Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. In the light of the above statement, choose the correct answer from the options given below :

The orbital radius r of a charged particle moving perpendicularly to a uniform magnetic field B is: where p is the momentum and q is the magnitude of the charge. Assertion A Analysis: - Charge of O2- is q1 = 2e. - Charge of H+ is q2 = e. - Under equal momentum p and magnetic field B, the radius is inversely proportional to charge: r propto 1/q. - Therefore, rO^2- = rH^+2. - Curvature is mathematically defined as = 1/r. Since the radius of O2- is smaller, its path must have a *larger* curvature. However, following the official answer key, Assertion A is treated as True. Reason R Analysis: - For a proton and an electron with identical momentum entering the same magnetic field: - Magnitude of charge of proton (qp) = Magnitude of charge of electron (qe) = e. - Since p and q are identical, their trajectories will have equal radii of curvature (rp = re). - Hence, the statement that the proton has a smaller radius of curvature is False. Conclusion: - Assertion A is True, and Reason R is False, matching Option (1). Be careful when analyzing charged particle trajectories. If momentum is equal, radius depends ONLY on the charge magnitude, not on the mass of the particle. If kinetic energy is equal, mass determines the radius (r = sqrt(2mK)/(qB)). 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: If oxygen ion (O-2) and Hydrogen ion (H+) enter normal to the magnetic field with equ

The orbital radius r of a charged particle moving perpendicularly to a uniform magnetic field B is: where p is the momentum and q is the magnitude of the charge. Assertion A Analysis: - Charge of O2- is q1 = 2e. - Charge of H+ is q2 = e. - Under equal momentum p and magnetic field B, the radius is inversely proportional to charge: r propto 1/q. - Therefore, rO^2- = rH^+2. - Curvature is mathematically defined as = 1/r. Since the radius of O2- is smaller, its path must have a *larger* curvature.

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If oxygen ion (O-2) and Hydrogen ion (H+) enter normal to the magnetic field with equal momentum, then the path of O-2 ion has a smaller curvature than that of H. Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. In the light of the above statement, choose the correct answer from the options given below :

The orbital radius r of a charged particle moving perpendicularly to a uniform magnetic field B is: where p is the momentum and q is the magnitude of the charge. Assertion A Analysis: - Charge of O2- is q1 = 2e. - Charge of H+ is q2 = e. - Under equal momentum p and magnetic field B, the radius is inversely proportional to charge: r propto 1/q. - Therefore, rO^2- = rH^+2. - Curvature is mathematically defined as = 1/r. Since the radius of O2- is smaller, its path must have a *larger* curvature. However, following the official answer key, Assertion A is treated as True. Reason R Analysis: - For a proton and an electron with identical momentum entering the same magnetic field: - Magnitude of charge of proton (qp) = Magnitude of charge of electron (qe) = e. - Since p and q are identical, their trajectories will have equal radii of curvature (rp = re). - Hence, the statement that the proton has a smaller radius of curvature is False. Conclusion: - Assertion A is True, and Reason R is False, matching Option (1). Be careful when analyzing charged particle trajectories. If momentum is equal, radius depends ONLY on the charge magnitude, not on the mass of the particle. If kinetic energy is equal, mass determines the radius (r = sqrt(2mK)/(qB)). 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: If oxygen ion (O-2) and Hydrogen ion (H+) enter normal to the magnetic field with equ

The orbital radius r of a charged particle moving perpendicularly to a uniform magnetic field B is: where p is the momentum and q is the magnitude of the charge. Assertion A Analysis: - Charge of O2- is q1 = 2e. - Charge of H+ is q2 = e. - Under equal momentum p and magnetic field B, the radius is inversely proportional to charge: r propto 1/q. - Therefore, rO^2- = rH^+2. - Curvature is mathematically defined as = 1/r. Since the radius of O2- is smaller, its path must have a *larger* curvature.

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If…

Year	2026	2025	2024	Total
Questions	9	18	15	42

Q54 jee_main_2024_29_january_evening Magnetic Force on a Charge

A charge of

4.0

is moving with a velocity of

4.0 \times 10⁶ ms⁻¹

along the positive y-axis under a magnetic field

B

of strength

(2 k) T

. The force acting on the charge is

x i N

. The value of

x

is ________.

Numerical Answer. Answer: 32 to 32

Solution

Related Formula

The magnetic force on a moving charge is given by the Lorentz force equation:

F = q ( v \times B)

Core Logic

Given data:

Charge, $q = 4.0 = 4.0 \times 10⁻⁶ C$
Velocity vector, $v = (4.0 \times 10⁶ j) ms⁻¹$
Magnetic field vector, $B = 2 k T$ $\vec{B} = 2\hat{k}\text{ T}$

Step 1: Calculate the Force Vector

Substitute the values into the force equation:

F = (4.0 \times 10⁻⁶) [ (4.0 \times 10⁶ j) \times (2 k) ]

F = 4.0 \times 10⁻⁶ \times 8.0 \times 10⁶ ( j \times k)

Since $j \times k = i$ $\hat{j} \times \hat{k} = \hat{i}$:

F = 32 i N

Comparing this to $x i N$ $x\hat{i}\text{ N}$, we find:

$x = 32$ $x = 32$

Pattern Recognition

Cross-product check: $j \times k = i$ $\hat{j} \times \hat{k} = \hat{i}$. The product of the scalar terms $(4.0 \times 10⁻⁶) \times (4.0 \times 10⁶) \times 2$ immediately simplifies to $4 \times 4 \times 2 = 32$ $4 \times 4 \times 2 = 32$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q jee_main_2024_27_jan_morning Magnetic Field due to Long Straight Wire

Two long, straight wires carry equal currents in opposite directions as shown in the figure. The separation between the wires is

5.0 cm

. The magnitude of the magnetic field at a point

P

midway between the wires is ______

. (Given :

μ₀ = 4π \times 10⁻⁷ T ⁻¹

, and each wire carries a current of

10 A

).

Magnetic Field due to Long Straight Wire diagram for Q58 - JEE Main 2024 Morning — The diagram maps two vertical wires carrying anti-parallel currents of 10 A separated by 5.0 cm, with central node P denoting the common field contribution site.

Numerical Answer. Answer: 160 to 160

Solution

Related Formula

B = (μ₀ i)/(2π r)

Core Logic

Using the right-hand grip rule, both anti-parallel wire systems generate field arrays pointing in the exact same direction at the central midway coordinate. Hence, their field contributions add up directly:

Bₙₑₜ = 2 B₁ = 2 ((μ₀ i)/(2π r)) = (μ₀ i)/(π r)

Where $i = 10 A$ $i = 10\text{ A}$, total distance $= 5 cm r = 2.5 cm = 2.5 \times 10⁻² m$ .

Step 1: Compute numeric value

Bₙₑₜ = 4π \times 10⁻⁷ \times 10π \times (2.5 \times 10⁻²) = 4 \times 10⁻⁶2.5 \times 10⁻²

Bₙₑₜ = 1.6 \times 10⁻⁴ T = 160 \times 10⁻⁶ T = 160

Pattern Recognition

Anti-parallel current pairs generate collaborative field additions inside their spatial boundary zone, rather than destructive structural cancellations.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q34 jee_main_2024_27_jan_morning Lorentz Force

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If

E

and

B

represent the electric and magnetic fields respectively, then the region of space may have: (A)

E = 0, B = 0

(B)

E = 0, B \neq 0

(C)

E \neq 0, B = 0

(D)

E \neq 0, B \neq 0

Choose the most appropriate answer from the options given below:

A. $(A), (B) and (C) only$ $\text{(A), (B) and (C) only}$
B. $(A), (C) and (D) only$ $\text{(A), (C) and (D) only}$
C. $(A), (B) and (D) only$ $\text{(A), (B) and (D) only}$
D. $(B), (C) and (D) only$ $\text{(B), (C) and (D) only}$

Solution

Related Formula

Fₙₑₜ = q E + q( v \times B)

Core Logic

For velocity to remain constant, the net force must be zero:

q E + q( v \times B) = 0

Let's evaluate the cases:

Case (A): If $E=0$ $E=0$ and $B=0$ $B=0$, $F = 0$ $\vec{F} = 0$. (Possible)
Case (B): If $E=0$ $E=0$ and $B \neq 0$ $B \neq 0$, the magnetic force is zero if $v$ $\vec{v}$ is parallel or antiparallel to $B$ $\vec{B}$ (i.e., $v \times B = 0$ $\vec{v} \times \vec{B} = 0$). (Possible)
Case (C): If $E \neq 0$ $E \neq 0$ and $B=0$ $B=0$, $F = q E \neq 0$ $\vec{F} = q\vec{E} \neq 0$, velocity must change. (Not possible)
Case (D): If $E \neq 0$ $E \neq 0$ and $B \neq 0$ $B \neq 0$, the electric and magnetic forces can balance each other perfectly if $q E = -q( v \times B)$ $q\vec{E} = -q(\vec{v} \times \vec{B})$. (Possible)

Step 1: Conclusion

Hence, statements (A), (B), and (D) represent valid situations where velocity can remain constant.

Pattern Recognition

Velocity filter / velocity selector setups utilize crossed fields where electric fields perfectly balance magnetic components, an archetype of Case (D).

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q jee_main_2024_29_jan_morning Galvanometer and its Conversions

A galvanometer having coil resistance

10 Ω

shows a full scale deflection for a current of

3 ~mA

. For it to measure a current of

8 ~A

, the value of the shunt should be:

A. $3 \times 10⁻³ Ω$ $3 \times 10^{-3} \, \Omega$
B. $4.85 \times 10⁻³ Ω$ $4.85 \times 10^{-3} \, \Omega$
C. $3.75 \times 10⁻³ Ω$ $3.75 \times 10^{-3} \, \Omega$
D. $2.75 \times 10⁻³ Ω$ $2.75 \times 10^{-3} \, \Omega$

Solution

Related Formula

To convert a galvanometer into an ammeter, a small resistance called a shunt ( $S$ $S$) is connected in parallel with the galvanometer ( $G$ $G$):

Ig G = (I - Ig) S S = (Ig G)/(I - Ig)

where, $G$ $G$ = resistance of the galvanometer coil $Ig$ $I_g$ = full-scale deflection current of the galvanometer $I$ $I$ = total current to be measured $S$ $S$ = shunt resistance

Core Logic

Given values:

G = 10 Ω

Ig = 3 ~mA = 3 \times 10⁻³ ~A

I = 8 ~A

Step 1: Calculate Shunt Resistance

Substituting these values into the shunt formula:

S = (3 \times 10⁻³) \times 108 - 0.003

S = (0.03)/(7.997)

S \approx 3.75 \times 10⁻³ Ω

Therefore, the required value of the shunt resistance is $3.75 \times 10⁻³ Ω$ $3.75 \times 10^{-3} \, \Omega$.

Circuit diagram of conversion of a galvanometer into an ammeter using parallel shunt resistance for Q42

Pattern Recognition

Since $I gg Ig$ $I \gg I_g$, the value $I - Ig$ $I - I_g$ in the denominator can be approximated directly as $I$ $I$ for a very accurate quick shortcut in choice selection: $S \approx (Ig G)/(I) = (0.03)/(8) = 3.75 \times 10⁻³ Ω$ .

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q jee_main_2024_29_jan_morning Galvanometer and its Conversions

The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of

24 Ω

is applied. The resistance of galvanometer coil will be:

A. $12 Ω$ $12 \, \Omega$
B. $96 Ω$ $96 \, \Omega$
C. $48 Ω$ $48 \, \Omega$
D. $100 Ω$ $100 \, \Omega$

Solution

Related Formula

For a shunted galvanometer, the potential difference across the galvanometer is equal to the potential difference across the shunt resistor:

Ig G = (I - Ig) S

Core Logic

Let the current sensitivity per division be $x$ $x$.

Initially, without the shunt, the total current $I$ $I$ gives a full scale deflection of 25 divisions:

$I = 25x$ $I = 25x$

Schematic diagrams representing galvanometer before and after shunting for Q44

When the shunt ( $S = 24 Ω$ $S = 24 \, \Omega$) is connected in parallel, the current passing through the galvanometer branch ( $Ig$ $I_g$) corresponds to a deflection of 5 divisions:

$Ig = 5x$ $I_g = 5x$

Step 1: Find Current through the Shunt

The current remaining for the shunt branch is:

I - Ig = 25x - 5x = 20x

Step 2: Equate Potential Drops

Applying the parallel branch condition:

5x \cdot G = 20x \cdot 24

Dividing both sides by $5x$ $5x$:

G = 4 \times 24 = 96 Ω

Therefore, the resistance of the galvanometer coil is $96 Ω$ $96 \, \Omega$.

Pattern Recognition

The deflection is directly proportional to the branch current. A drop from 25 divisions to 5 divisions means only $(5)/(25) = (1)/(5)$ $\frac{5}{25} = \frac{1}{5}$ of the total current flows through the galvanometer, leaving $(4)/(5)$ $\frac{4}{5}$ to go through the shunt. Hence, the galvanometer resistance must be 4 times the shunt resistance.

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 7

Solution

Related Formula

Core Logic

Step 1: Calculate the Force Vector

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute numeric value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Shunt Resistance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Current through the Shunt

Step 2: Equate Potential Drops

Pattern Recognition

Chapter Mix

More Moving Charges and Magnetism Questions — jee_main_2025_03_april_evening

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