Subtopic: Biot-Savart Law & Ampere's Circuital Law
Question Statement:
Cyclotrons are used to accelerate ions like deuterons (d) and \alpha particles. Keeping
the magnetic field same for both, d and \alpha are extracted with energies 10 MeV and
20 MeV with extraction radii rd and r\alpha, respectively. Taking the masses
Md = 2000 MeV/c2 and M\alpha = 4000 MeV/c2, t he value of
r\alpha
rd
(in integer) is
______
Official Answer Key:
1 to 1
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.59)**
**1. Cyclotron Acceleration Physics**:
- The non-relativistic kinetic energy of an accelerated ion of mass $m$ and charge $q$ extracted at radius $r$ in magnetic field $B$ is:
$E = \frac{q^2 B^2 r^2}{2m} \implies r^2 = \frac{2m E}{q^2 B^2}$
**2. Step-by-Step Ratio Derivation**:
- Given parameters:
- Constant magnetic field $B$.
- Deuteron ($d$): $m_d = 2m_p$, charge $q_d = e$, extracted energy $E_d = 10\text{ MeV}$.
- Alpha particle ($\alpha$): $m_\alpha = 4m_p$, charge $q_\alpha = 2e$, extracted energy $E_\alpha = 20\text{ MeV}$.
- Compute extraction radius squared for each:
$r_d^2 = \frac{2(2m_p)(10\text{ MeV})}{e^2 B^2} = \frac{40 m_p}{e^2 B^2}$
$r_\alpha^2 = \frac{2(4m_p)(20\text{ MeV})}{(2e)^2 B^2} = \frac{160 m_p}{4e^2 B^2} = \frac{40 m_p}{e^2 B^2}$
- Taking the ratio:
$\frac{r_d^2}{r_\alpha^2} = \frac{40}{40} = 1 \implies \frac{r_d}{r_\alpha} = 1$
**3. Official Verification**:
The verified official IIT answer is **1** (Numerical range: `1 to 1`).
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