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…

Q51 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

4pisqrt3mathrm~A

is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be

x times 10^-7mathrm~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

at distance

r

:

B_1 = fracmu_0 I4pi r(sintheta_1 + sintheta_2)

Total field for a regular hexagon (

n=6

):

B = 6 times B_1

### Core Logic Total perimeter

= 6 cdot a = 4pi implies textside length a = frac4pi6 = frac2pi3mathrm~m

. For a regular hexagon segment, the interior angles relative to the normal vector are

theta_1 = theta_2 = 30^circ

. The normal distance

r

from the center to a side is:

r = fraca2 cot(30^circ) = frac4pi2 times 6 times sqrt3 = fracsqrt3pi3 = fracpisqrt3mathrm~m

### Step 1: Calculate Total Magnetic Field Substitute

r

and

I = 4pisqrt3mathrm~A

into the hexagon configuration:

B = 6 times left[ fracmu_0 I4pi r (sin(30^circ) + sin(30^circ)) right]

B = 6 times left[ frac10^-7 times 4pisqrt3left(fracsqrt3pi3right) times (0.5 + 0.5) right]

B = 6 times left[ frac10^-7 times 4pisqrt3 times 3sqrt3pi times 1 right]

B = 6 times [ 4 times 3 times 10^-7 ] = 6 times 12 times 10^-7 = 72 times 10^-7mathrm~T

Thus,

x = 72

. ### Pattern Recognition For regular polygons, the normal distance

r

to the side is always

r = fraca2cot(fracpin)

. The contribution from all

n

symmetric segments adds up constructively. ### Evaluation Rubric / Model Answer null ### 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

R_1

and

R_2

respectively. The mass ratio of X and Y is:

A. $left(fracR_2R_1right)^2$ $\left(\frac{R_2}{R_1}\right)^2$
B. $left(fracR_1R_2right)^2$ $\left(\frac{R_1}{R_2}\right)^2$
C. $left(fracR_1R_2right)$ $\left(\frac{R_1}{R_2}\right)$
D. $left(fracR_2R_1right)$ $\left(\frac{R_2}{R_1}\right)$

Solution

### Related Formula The radius

R

of the path of a charged particle moving perpendicular to a magnetic field

B

is:

R = fracmvqB = fracpqB

In terms of kinetic energy

K

:

R = fracsqrt2mKqB

Since the particle is accelerated through potential

V

, kinetic energy

K = qV

:

R = fracsqrt2mqVqB

R = frac1B sqrtfrac2mVq

### Core Logic For both particles X and Y, the following parameters are the same: * Potential Difference,

V

* Magnetic Field,

B

* Charge,

q

Therefore, we have the proportionality:

R propto sqrtm implies R^2 propto m

### Step 1: Calculate Mass Ratio Using the proportionality relationship:

fracm_1m_2 = left( fracR_1R_2 right)^2

Thus, the mass ratio of X and Y is

left(fracR_1R_2right)^2

. ### Pattern Recognition Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as

R propto sqrtm

. Squaring both sides yields

m propto R^2

instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Q54 jee_main_2024_29_january_evening Magnetic Force on a Charge

A charge of

4.0\ mutextC

is moving with a velocity of

4.0 times 10^6text ms^-1

along the positive y-axis under a magnetic field

vecB

of strength

(2hatk)text T

. The force acting on the charge is

xhatitext 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:

vecF = q (vecv times vecB)

### Core Logic Given data: * Charge,

q = 4.0\ mutextC = 4.0 times 10^-6text C

* Velocity vector,

vecv = (4.0 times 10^6hatj)text ms^-1

* Magnetic field vector,

vecB = 2hatktext T

### Step 1: Calculate the Force Vector Substitute the values into the force equation:

vecF = (4.0 times 10^-6) left[ (4.0 times 10^6hatj) times (2hatk) right]

vecF = 4.0 times 10^-6 times 8.0 times 10^6 (hatj times hatk)

Since

hatj times hatk = hati

:

vecF = 32hatitext N

Comparing this to

xhatitext N

, we find:

x = 32

### Pattern Recognition Cross-product check:

hatj times hatk = hati

. The product of the scalar terms

(4.0 times 10^-6) times (4.0 times 10^6) times 2

immediately simplifies to

4 times 4 times 2 = 32

. ### Evaluation Rubric / Model Answer null ### 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

vecE

and

vecB

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

Solution

### Related Formula

vecF_textnet = qvecE + q(vecv times vecB)

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

qvecE + q(vecv times vecB) = 0

Let's evaluate the cases: - Case (A): If

E=0

and

B=0

,

vecF = 0

. (Possible) - Case (B): If

E=0

and

B neq 0

, the magnetic force is zero if

vecv

is parallel or antiparallel to

vecB

(i.e.,

vecv times vecB = 0

). (Possible) - Case (C): If

E neq 0

and

B=0

,

vecF = qvecE neq 0

, velocity must change. (Not possible) - Case (D): If

E neq 0

and

B neq 0

, the electric and magnetic forces can balance each other perfectly if

qvecE = -q(vecv times vecB)

. (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). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Q58 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

5.0text cm

. The magnitude of the magnetic field at a point

P

midway between the wires is ______

mutextT

. (Given :

mu_0 = 4pi times 10^-7text TcdottextmcdottextA^-1

, and each wire carries a current of

10text 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 = fracmu_0 i2pi 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:

B_textnet = 2 B_1 = 2 left(fracmu_0 i2pi rright) = fracmu_0 ipi r

Where

i = 10text A

, total distance

= 5text cm implies r = 2.5text cm = 2.5 times 10^-2text m

. ### Step 1: Compute numeric value

B_textnet = frac4pi times 10^-7 times 10pi times (2.5 times 10^-2) = frac4 times 10^-62.5 times 10^-2

B_textnet = 1.6 times 10^-4text T = 160 times 10^-6text T = 160\ mutextT

### Pattern Recognition Anti-parallel current pairs generate collaborative field additions inside their spatial boundary zone, rather than destructive structural cancellations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Solution & Explanation

Reference Study Guides

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

Solution

Solution

Solution

Solution

Solution

More Moving Charges and Magnetism Questions — jee_main_2025_03_april_evening

We Map Every Repeating Question in Competitive Exams.

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

Solution & Explanation

Reference Study Guides

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

Solution

Solution

Solution

Solution

Solution

More Moving Charges and Magnetism Questions — jee_main_2025_03_april_evening

We Map Every Repeating Question in Competitive Exams.

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