Radius vs Energy in Magnetic Field — JEE Concept in 59 Sec 🔥

❓ Question

A particle of charge $q$, mass $m$ and kinetic energy $E$ enters a uniform magnetic field perpendicular to its velocity and undergoes a circular arc of radius $r$.

Which of the following curves correctly represents the variation of $r$ with $E$?

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🖼️ Question Image

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✍️ Short Explanation

This is a pure relation-based magnetism question.
No numbers. No tricks.
👉 Sirf radius–energy dependence samajhni hai.

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🔹 Step 1 — Magnetic force causes circular motion

For a charged particle moving perpendicular to magnetic field $B$:

$$qvB=\frac{m{v}^2}{r}$$

So radius of circular path:

$$\boxed{{\displaystyle r=\frac{mv}{qB}}}$$

📌 Radius speed $v$ par depend karta hai.

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🔹 Step 2 — Connect speed with kinetic energy

Kinetic energy:

$$E=\frac{1}{2}m{v}^2$$

So:

$$v=\sqrt{\frac{2E}{m}}$$

Substitute in radius expression:

$$r=\frac{m}{qB}\sqrt{\frac{2E}{m}}$$

Hence:

$$\boxed{r\propto \sqrt{E}}$$

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🔹 Step 3 — Nature of r vs E graph (MOST IMPORTANT 🔥)**

From:

$$r\propto\sqrt{E}$$

We conclude:

• $r$ increases with $E$
  

• Increase is not linear

• Slope decreases gradually as $E$ increases

👉 This gives a curve that:

• Starts from origin

• Rises quickly initially

• Then flattens slowly

📌 Concave downward curve ✔️

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🔹 Step 4 — Eliminate wrong graph options

Let’s match with given curves:

❌ Straight line → implies $r\propto E$ (wrong)
❌ Upward bending curve → implies $r\propto E^2$ (wrong)
❌ Decreasing curve → physically impossible
✔️ Increasing but flattening curve → $r\propto\sqrt{E}$ (correct)

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🔹 Step 5 — Physical intuition (JEE check 🧠)**

• Magnetic field does no work

• Higher energy ⇒ higher speed

• Faster particle is harder to bend

• Hence radius becomes larger, but slowly

🧠 One-line intuition:

Energy badhi ⇒ bending kam effective ⇒ radius bada

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✅ Final Answer

✔️ The correct curve is the one where $r$ increases with $E$ as a square-root function
(increasing, concave downward curve).

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⭐ Golden Summary Box

• Magnetic field → circular motion

• $r=\dfrac{mv}{qB}$

• $v\propto\sqrt{E}$

• $r\propto\sqrt{E}$

• Graph: rising, concave down