back to directory
NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

When a proton is released from rest in a room, it starts with an initial acceleration a0a_0 towards the east. When it is projected towards the north with a speed v0v_0, it moves with initial acceleration 3a03a_0 towards east. The electric and magnetic fields in the room are:

A

ma0e\frac{ma_0}{e} west, ma0ev0\frac{ma_0}{ev_0} up

B

ma0e\frac{ma_0}{e} west, 2ma0ev0\frac{2ma_0}{ev_0} down

C

ma0e\frac{ma_0}{e} east, 2ma0ev0\frac{2ma_0}{ev_0} down

D

ma0e\frac{ma_0}{e} east, 2ma0ev0\frac{2ma_0}{ev_0} up

Step-by-Step Solution

  1. Analyze Electric Force (Case 1): When released from rest (v=0v=0), the magnetic force Fm=q(v⃗×B⃗)F_m = q(\vec{v} \times \vec{B}) is zero. The only force acting is the electric force Fe=eEF_e = eE. Fe=ma0F_e = ma_0 (towards East). eE=ma0  ⟹  E=ma0eeE = ma_0 \implies E = \frac{ma_0}{e}.
  • Since the proton (positive charge) accelerates East, the Electric Field E⃗\vec{E} is directed towards East .
  1. Analyze Magnetic Force (Case 2): When projected North with speed v0v_0, the total initial acceleration is 3a03a_0 towards East. Net Force F⃗net=m(3a0)\vec{F}_{net} = m(3a_0) (East). F⃗net=F⃗e+F⃗m\vec{F}_{net} = \vec{F}_e + \vec{F}_m.
  • 3ma0=ma0+Fm  ⟹  Fm=2ma03ma_0 = ma_0 + F_m \implies F_m = 2ma_0 (directed towards East).
  1. Determine Magnetic Field Direction: The magnetic force is given by F⃗m=e(v⃗×B⃗)\vec{F}_m = e(\vec{v} \times \vec{B}) . We have v⃗\vec{v} towards North and F⃗m\vec{F}_m towards East. Using the Right-Hand Rule (or vector product): North ×\times B⃗\vec{B} = East.
  • (j^×k^=i^)(\hat{j} \times \hat{k} = \hat{i}). Thus, North ×\times Up = East. The magnetic field must be directed Upwards .
  1. Calculate Magnitude: ev0Bsin⁡(90∘)=2ma0e v_0 B \sin(90^\circ) = 2ma_0 B=2ma0ev0B = \frac{2ma_0}{ev_0}.
  2. Conclusion: Electric field is ma0e\frac{ma_0}{e} East, Magnetic field is 2ma0ev0\frac{2ma_0}{ev_0} Up.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMprotonreleasedstartsinitialacceleration

More MOVING CHARGES AND MAGNETISM Questions

View all

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron:

A.will turn towards right of direction of motion
B.will turn towards left of direction of motion
C.speed will decrease
D.speed will increase
EasySolve

A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{-4} \text{ m}^2$ carries a current of $2.0 \text{ A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} \text{ T}$ making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be

A.$3 \times 10^{-3} \text{ N m}$
B.$1.5 \times 10^{-3} \text{ N m}$
C.$1.5 \times 10^{-2} \text{ N m}$
D.$3 \times 10^{-2} \text{ N m}$
EasySolve

A metallic rod of mass per unit length of 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction of 0.25 T is acting on it in the vertical direction. What is the current flowing through the rod to keep it stationary?

A.7.14 A
B.5.98 A
C.14.76 A
D.11.32 A
MediumSolve

A voltmeter has a resistance of $G$ ohms and range $V$ volts. The value of resistance used in series to convert it into a voltmeter of range $nV$ volts is:

A.$nG$
B.$(n-1)G$
C.$\frac{G}{n}$
D.$\frac{G}{n-1}$
MediumSolve

The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is:

A.40 Ω
B.25 Ω
C.250 Ω
D.500 Ω
EasySolve

A long solenoid of radius $1 \text{ mm}$ has $100$ turns per mm. If $1 \text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:

A.$6.28 \times 10^{-4} \text{ T}$
B.$6.28 \times 10^{-2} \text{ T}$
C.$12.56 \times 10^{-2} \text{ T}$
D.$12.56 \times 10^{-4} \text{ T}$
EasySolve

A square loop ABCD carrying a current $i$ is placed near and coplanar with a long straight conductor XY carrying a current $I$. The net force on the loop will be:

A.$\frac{\mu_0 I i}{2\pi}$
B.$\frac{2\mu_0 I i L}{3\pi}$
C.$\frac{\mu_0 I i L}{2\pi}$
D.$\frac{2\mu_0 I i}{3\pi}$
HardSolve

An electron moving in a circular orbit of radius $r$ makes $n$ rotations per second. The magnetic field produced at the centre has magnitude:

A.$\frac{\mu_0 n e}{2\pi r}$
B.Zero
C.$\frac{n^2 e}{r}$
D.$\frac{\mu_0 n e}{2r}$
EasySolve

Why 32,000+ students choose TopperSquare

Everything you need to crack NEET

15,000+ Chapter MCQs

Physics, Chemistry, Biology — chapter-wise, topic-wise practice

AI Performance Analytics

Know your weak topics. Get a personalised study plan.

Full NEET Mock Tests

180 questions, timed, with detailed score analysis

PYQ Sets 2010–2024

15 years of past papers with solved explanations

Start Practising Free

No credit card · 30-second signup · Free forever plan included

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →

Explore more subjects:Physics MCQs•Chemistry MCQs•Biology MCQs•Botany MCQs•Zoology MCQs•NEET Mock Tests•NEET PYQs