Olympiad Wallah
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🎯 Olympiad & JEE Prep Channel Your go-to space for: 🔹 NSEC | NSEB | NSEP | NSEA | NSEJS | IOQM 🔹 Advanced Series: Physics, Chem, Bio, Maths 🔹 JEE Excellence & Test Series Stay updated. Stay prepared. Let’s crack it! 💥
显示更多📈 Telegram 频道 Olympiad Wallah 的分析概览
频道 Olympiad Wallah (@olympiad_wallah) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 15 534 名订阅者，在 教育 类别中位列第 12 983，并在 印度 地区排名第 26 892 位。
📊 受众指标与增长动态
自 невідомо 创建以来，项目保持高速增长，吸引了 15 534 名订阅者。
根据 24 七月, 2026 的最新数据，频道保持稳定运转。过去 30 天订阅人数变化为 402，过去 24 小时变化为 5，整体触达仍然可观。
- 认证状态： 未认证
- 互动率 (ER)： 平均受众互动率为 18.20%。内容发布后 24 小时内通常能获得 6.23% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 823 次浏览，首日通常累积 966 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 8。
- 主题关注点： 内容集中在 champ, revision, aspirant, aaj, olympiadwallah 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台：
“🎯 Olympiad & JEE Prep Channel
Your go-to space for:
🔹 NSEC | NSEB | NSEP | NSEA | NSEJS | IOQM
🔹 Advanced Series: Physics, Chem, Bio, Maths
🔹 JEE Excellence & Test Series
Stay updated. Stay prepared. Let’s crack it! 💥”
凭借高频更新（最新数据采集于 25 七月, 2026），频道始终保持新鲜度与高覆盖。分析显示受众积极互动，使其成为 教育 类别中的关键影响点。
15 534
订阅者
+524 小时
+467 天
+40230 天
数据加载中...
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| 日期 | 订阅者增长 | 提及 | 频道 | |
| 25 七月 | +22 | |||
| 24 七月 | +5 | |||
| 23 七月 | +1 | |||
| 22 七月 | +14 | |||
| 21 七月 | +5 | |||
| 20 七月 | +18 | |||
| 19 七月 | +7 | |||
| 18 七月 | +8 | |||
| 17 七月 | +12 | |||
| 16 七月 | +4 | |||
| 15 七月 | +25 | |||
| 14 七月 | +5 | |||
| 13 七月 | +11 | |||
| 12 七月 | +26 | |||
| 11 七月 | +13 | |||
| 10 七月 | +15 | |||
| 09 七月 | +9 | |||
| 08 七月 | +3 | |||
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| 06 七月 | +9 | |||
| 05 七月 | +14 | |||
| 04 七月 | +45 | |||
| 03 七月 | +17 | |||
| 02 七月 | +33 | |||
| 01 七月 | +9 |
频道帖子
A proton with kinetic energy 1 MeV moves from south to north. It experiences an acceleration of 10¹² m/s² in a magnetic field directed from west to east. The magnitude of the magnetic field is:
| 2 | A proton enters a uniform magnetic field parallel to the field lines. The magnetic force acting on the proton is: | 455 |
| 3 |  📒 MODULAR AIRTHEMETIC in One Page!
Save this quick revision sheet and strengthen one of the most important topic for Olympiads. Perfect for last-minute revision! 🚀 | 725 |
| 4 | ⚡ FORCE ON A MOVING CHARGE — QUICK NOTES
🔹 Magnetic Force on a Moving Charge
A charged particle moving in a magnetic field experiences a force given by:
F = qvB sinθ
where,
F = Magnetic Force
q = Charge
v = Velocity of the Particle
B = Magnetic Field
θ = Angle between v and B
🔹 Special Cases
• θ = 0° or 180°
F = 0
(Motion parallel or antiparallel to the magnetic field)
• θ = 90°
F = qvB (Maximum)
(Motion perpendicular to the magnetic field)
🔹 Direction of Magnetic Force
Use Fleming's Left-Hand Rule (for current) or the Right-Hand Rule for a moving positive charge.
For a negative charge, the force is opposite to the direction obtained for a positive charge.
🔹 Circular Motion in a Magnetic Field
Radius of Circular Path:
r = mv/qB
🔹 Time Period
T = 2πm/qB
🔹 Cyclotron Frequency
f = qB/2πm
🔹 Angular Frequency
ω = qB/m
⚡ OLYMPIAD FACTS
✓ Magnetic force is always perpendicular to both the velocity and the magnetic field.
✓ A magnetic field changes only the direction of motion, not the speed of a charged particle.
✓ A stationary charge experiences no magnetic force.
✓ Magnetic force does no work on a charged particle.
✓ A charged particle moves in a circular path if its velocity is perpendicular to the magnetic field.
🎯 MUST REMEMBER
F = qvB sinθ
F_max = qvB
r = mv/qB
T = 2πm/qB
f = qB/2πm
ω = qB/m | 759 |
| 5 | A current I flows through an infinitely long thin-walled cylindrical conductor. The magnetic field at any point inside the conductor is: | 818 |
| 6 | A long straight wire of radius a carries a steady current I, uniformly distributed across its cross-section. The ratio of the magnetic field at a distance a/2 and 2a from the axis of the wire is | 824 |
| 7 | ⚡ AMPERE'S CIRCUITAL LAW — QUICK NOTES
🔹 Ampere's Circuital Law
The line integral of the magnetic field around any closed path is equal to μ₀ times the net current enclosed by the path.
Formula:
∮B·dl = μ₀Iₑₙc
where,
B = Magnetic Field
dl = Infinitesimal Length Element
Iₑₙc = Net Current Enclosed
μ₀ = Permeability of Free Space
🔹 Magnetic Field Due to a Long Straight Wire
B = μ₀I/2πr
🔹 Magnetic Field Inside a Long Solenoid
B = μ₀nI
where,
n = Number of Turns per Unit Length
🔹 Magnetic Field Inside a Toroid
B = μ₀NI/2πr
where,
N = Total Number of Turns
r = Mean Radius of the Toroid
🔹 Magnetic Field Outside a Long Solenoid
B ≈ 0
🔹 Magnetic Field Outside an Ideal Toroid
B = 0
⚡ OLYMPIAD FACTS
✓ Ampere's Circuital Law is applicable to highly symmetric current distributions.
✓ The magnetic field inside a long solenoid is nearly uniform.
✓ The magnetic field outside an ideal solenoid is approximately zero.
✓ The magnetic field outside an ideal toroid is zero.
✓ The direction of the magnetic field is determined by the Right-Hand Thumb Rule.
🎯 MUST REMEMBER
∮B·dl = μ₀Iₑₙc
B = μ₀I/2πr
B = μ₀nI
B = μ₀NI/2πr
B ≈ 0 (Outside Long Solenoid)
B = 0 (Outside Ideal Toroid) | 881 |
| 8 | 📘 Probability Theory Notes are here 🔥
From basic concepts to Olympiad level tricks — everything covered for quick revision 🚀 | 953 |
| 9 | The magnetic field at a distance d from a long straight wire carrying a current I is B. What will be the magnetic field at a distance 2d from the wire? | 1 168 |
| 10 | The magnetic field at the centre of a circular coil of radius R carrying current I is: | 1 121 |
| 11 | ⚡ BIOT–SAVART LAW — QUICK NOTES
🔹 Biot–Savart Law
The magnetic field due to a small current element is directly proportional to the current, the length of the current element, and sinθ, and inversely proportional to the square of the distance from the element.
Formula:
dB = (μ₀/4π) (I dl sinθ)/r²
where,
dB = Magnetic Field due to Current Element
I = Current
dl = Current Element
r = Distance from the Element
θ = Angle between dl and r
🔹 Magnetic Field Due to a Long Straight Wire
B = μ₀I/2πr
🔹 Magnetic Field at the Centre of a Circular Loop
B = μ₀I/2R
🔹 Magnetic Field on the Axis of a Circular Loop
B = μ₀IR²/[2(R² + x²)³ᐟ²]
where,
x = Distance from the Centre Along the Axis
🔹 Magnetic Field Due to N Turns
B = μ₀NI/2R
🔹 Magnetic Field Inside a Long Solenoid
B = μ₀nI
where,
n = Number of Turns per Unit Length
🔹 Magnetic Field Inside a Toroid
B = μ₀NI/2πr
⚡ OLYMPIAD FACTS
✓ Biot–Savart Law is used to calculate the magnetic field produced by a current-carrying conductor.
✓ The direction of the magnetic field is given by the Right-Hand Thumb Rule.
✓ The magnetic field is maximum at the centre of a circular loop.
✓ The magnetic field at the centre of a complete circular loop is twice that of a long straight wire at the same distance (R).
✓ Magnetic field is directly proportional to current and inversely proportional to distance.
🎯 MUST REMEMBER
dB = (μ₀/4π)(I dl sinθ)/r²
B = μ₀I/2πr
B = μ₀I/2R
B = μ₀NI/2R
B = μ₀nI
B = μ₀NI/2πr
These are the most frequently used Biot–Savart Law formulas in Olympiad, JEE, and NEET problems. | 1 143 |
| 12 |  📒 CONGRUENCES in One Page!
Save this quick revision sheet and strengthen one of the most important topic for Olympiads. Perfect for last-minute revision! 🚀 | 1 162 |
| 13 |  Be ready Champs 💪🏻 | 1 425 |
| 14 |  Olympiad Wallah7 hours ago
🚨 IOQM Registration Closing Soon! ⏳
Don't miss your chance to begin your Olympiad journey! Register for IOQM 2026 before the deadline and take your first step towards the IMO.
📌 Register Now | 1 658 |
| 15 | If a current is passed through a spring, then the spring will: | 1 345 |
| 16 | Two circular coils A and B have radii R and 2R, respectively. If currents I and 2I flow through them, the ratio of the magnetic fields at their centres (B_A : B_B) is: | 1 341 |
| 17 | ⚡ MAGNETIC EFFECTS OF CURRENT — QUICK NOTE
🔹 Oersted's Experiment
A current-carrying conductor produces a magnetic field around it.
🔹 Magnetic Field Due to a Long Straight Wire
B = μ₀I/2πr
where,
B = Magnetic Field
I = Current
r = Perpendicular Distance from the Wire
μ₀ = Permeability of Free Space
🔹 Force on a Moving Charge
F = qvB sinθ
🔹 Force on a Current-Carrying Conductor
F = BIL sinθ
🔹 Torque on a Current Loop
τ = NIAB sinθ
where,
N = Number of Turns
A = Area of the Loop
🔹 Magnetic Dipole Moment
M = NIA
🔹 Torque on a Magnetic Dipole
τ = MB sinθ
🔹 Radius of Circular Motion
r = mv/qB
🔹 Cyclotron Frequency
f = qB/2πm
🔹 Time Period of Circular Motion
T = 2πm/qB
⚡ OLYMPIAD FACTS
✓ Magnetic field lines form closed loops.
✓ Magnetic force is always perpendicular to both velocity and magnetic field.
✓ A magnetic field does no work on a moving charge.
✓ A stationary charge experiences no magnetic force.
✓ The direction of the magnetic field around a straight wire is given by the Right-Hand Thumb Rule.
🎯 MUST REMEMBER
B = μ₀I/2πr
F = qvB sinθ
F = BIL sinθ
τ = NIAB sinθ
M = NIA
r = mv/qB
f = qB/2πm
T = 2πm/qB | 1 325 |
| 18 | 📘 Finite Series And Complex Numbers Notes are here 🔥
From basic concepts to Olympiad level tricks — everything covered for quick revision 🚀 | 1 317 |
| 19 | ⚡ KIRCHHOFF'S LAWS — QUICK NOTES
🔹 Kirchhoff's Current Law (KCL)
Statement:
The algebraic sum of currents at any junction is zero.
Formula:
ΣI = 0
or
ΣI_in = ΣI_out
🔹 Kirchhoff's Voltage Law (KVL)
Statement:
The algebraic sum of all potential differences (voltage rises and drops) around any closed loop is zero.
Formula:
ΣV = 0
🔹 Ohm's Law
V = IR
(Used with KVL to solve circuit problems.)
🔹 Voltage Drop Across a Resistor
V = IR
🔹 Internal Resistance of a Cell
V = E − Ir
where,
E = EMF of the Cell
I = Current
r = Internal Resistance
🔹 Cells in Series
E_eq = E₁ + E₂ + ...
r_eq = r₁ + r₂ + ...
🔹 Cells in Parallel (Identical Cells)
E_eq = E
r_eq = r/n | 1 541 |
| 20 | A 220 V, 1000 W electric bulb is connected to a 110 V supply. The power consumed by the bulb will be: | 1 477 |
