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 15 507 名订阅者，在 教育 类别中位列第 12 996，并在 印度 地区排名第 26 858 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 15 507 名订阅者。
根据 22 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 440，过去 24 小时变化为 14，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 17.78%。内容发布后 24 小时内通常能获得 6.25% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 757 次浏览，首日通常累积 970 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 23 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。
15 507
订阅者
+1424 小时
+607 天
+44030 天
数据加载中...
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| 日期 | 订阅者增长 | 提及 | 频道 | |
| 23 七月 | +1 | |||
| 22 七月 | +14 | |||
| 21 七月 | +5 | |||
| 20 七月 | +18 | |||
| 19 七月 | +7 | |||
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| 12 七月 | +26 | |||
| 11 七月 | +13 | |||
| 10 七月 | +15 | |||
| 09 七月 | +9 | |||
| 08 七月 | +3 | |||
| 07 七月 | +10 | |||
| 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
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?
| 2 | The magnetic field at the centre of a circular coil of radius R carrying current I is: | 354 |
| 3 | ⚡ 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. | 478 |
| 4 |  📒 CONGRUENCES in One Page!
Save this quick revision sheet and strengthen one of the most important topic for Olympiads. Perfect for last-minute revision! 🚀 | 696 |
| 5 |  Be ready Champs 💪🏻 | 927 |
| 6 |  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 | 1 198 |
| 7 | If a current is passed through a spring, then the spring will: | 1 020 |
| 8 | 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: | 977 |
| 9 | ⚡ 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 | 994 |
| 10 | 📘 Finite Series And Complex Numbers Notes are here 🔥
From basic concepts to Olympiad level tricks — everything covered for quick revision 🚀 | 1 009 |
| 11 | ⚡ 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 | 1 148 |
| 12 | A 220 V, 1000 W electric bulb is connected to a 110 V supply. The power consumed by the bulb will be: | 1 126 |
| 13 | Consider the following statements:
(A) Kirchhoff's Junction Law follows from the conservation of charge. (B) Kirchhoff's Loop Law follows from the conservation of energy. Choose the correct option: | 1 137 |
| 14 |  📒 FERMAT'S LITTLE THEORAM in One Page!
Save this quick revision sheet and strengthen one of the most important topic for Olympiads. Perfect for last-minute revision! 🚀 | 1 305 |
| 15 | ⚡ CAPACITORS — FORMULA SHEET
🔹 Capacitance
C = Q/V
where,
C = Capacitance
Q = Charge
V = Potential Difference
🔹 Parallel Plate Capacitor
C = ε₀A/d
where,
ε₀ = Permittivity of Free Space
A = Area of Each Plate
d = Separation Between Plates
🔹 Capacitor with Dielectric
C = Kε₀A/d
where,
K = Dielectric Constant
🔹 Charge Stored
Q = CV
🔹 Potential Difference
V = Q/C
🔹 Energy Stored in a Capacitor
U = ½CV²
U = ½QV
U = Q²/(2C)
🔹 Energy Density
u = ½εE²
where,
ε = Permittivity of the Medium
E = Electric Field
🔹 Electric Field Between Plates
E = V/d
🔹 Force Between Capacitor Plates
F = ½εAE²
🔹 Capacitance in Series
1/Ceq = 1/C₁ + 1/C₂ + 1/C₃ + ...
✓ Charge remains the same on each capacitor.
✓ Voltage is divided among the capacitors.
🔹 Capacitance in Parallel
Ceq = C₁ + C₂ + C₃ + ...
✓ Voltage remains the same across each capacitor.
✓ Charge is divided among the capacitors.
🔹 Dielectric Effect
New Capacitance:
C' = KC | 1 374 |
| 16 | Two capacitors of capacitances 3 μF and 6 μF are connected in series. The equivalent capacitance is | 1 402 |
| 17 | A capacitor is connected to a battery. A dielectric slab is completely inserted between its plates without disconnecting the battery. Which one of the following increases? | 1 396 |
| 18 | 📘 Elementary Combinatorics Notes are here 🔥
From basic concepts to Olympiad level tricks — everything covered for quick revision 🚀 | 1 498 |
| 19 | ⚡ ELECTRIC POTENTIAL & POTENTIAL ENERGY — QUICK NOTES
🔹 Electric Potential (V)
Electric potential at a point is the work done per unit positive test charge in bringing it from infinity to that point.
Formula:
V = W/q
Unit: Volt (V)
🔹 Potential Due to a Point Charge
V = kQ/r
where,
k = 1/4πϵ₀
Q = Source Charge
r = Distance from the Charge
🔹 Superposition Principle
The net electric potential at a point is the algebraic sum of the potentials due to all individual charges.
V_net = V₁ + V₂ + V₃ + ...
🔹 Potential Difference
ΔV = W/q
🔹 Electric Potential Energy
The energy possessed by a system of charges due to their positions.
🔹 Potential Energy of Two Point Charges
U = k(q₁q₂)/r
🔹 Potential Energy of a System of Charges
U = Σ k(qᵢqⱼ)/rᵢⱼ
(Sum over all distinct pairs of charges)
🔹 Relation Between Electric Field & Potential
E = −dV/dr
For a uniform electric field:
ΔV = −Ed
🔹 Equipotential Surface
✓ Electric potential is the same at every point.
✓ No work is done in moving a charge along an equipotential surface.
⚡ OLYMPIAD FACTS
✓ Electric potential is a scalar quantity.
✓ Electric potential may be positive, negative, or zero.
✓ Electric field is always perpendicular to an equipotential surface.
✓ Work done in moving a charge around a closed path in an electrostatic field is zero.
✓ At infinity, the electric potential due to an isolated point charge is taken as zero.
🎯 MUST REMEMBER
V = W/q
V = kQ/r
ΔV = W/q
U = k(q₁q₂)/r
V_net = ΣV
E = −dV/dr
ΔV = −Ed (Uniform Electric Field)
These are the most frequently used Electric Potential and Potential Energy concepts in Olympiad, JEE, and NEET problems. | 1 994 |
| 20 | 没有文字... | 2 107 |
