Question

A be am of ion with 2×100000 m/s enters normally into a uniform magnetic field of 4/100 T . If the the specific charge of the ions is 5 ×10000000 C/kg , the radius of the circular path described will be (1) 0.10m (2) 0.16 m. (3) 0.20m (4) 0.25 m.

Answer 1

Question #75075, Physics / Electromagnetism

A be am of ion with $2 \times 100000 \, \mathrm{m/s}$ enters normally into a uniform magnetic field of $4/100 \, \mathrm{T}$. If the specific charge of the ions is $5 \times 10000000 \, \mathrm{C/kg}$, the radius of the circular path described will be (1) $0.10 \, \mathrm{m}$ (2) $0.16 \, \mathrm{m}$ (3) $0.20 \, \mathrm{m}$ (4) $0.25 \, \mathrm{m}$.

Need to find:

**R-?**

$$\begin{array}{l} v = 2 \times 10^{5} \, \mathrm{m/s} \\ B = \frac{4}{100} \, \mathrm{T} \\ \frac{q}{m} = 5 \times 10^{7} \, \mathrm{C/kg} \\ \end{array}$$

Solution:

Lorentz force – (in Picture) $F_L = F_c$, where $F_L = qvB$, and $F_c = m \frac{v^2}{R}$.

$$\begin{array}{l} m \frac{v^2}{R} = qvB \rightarrow m \frac{v}{R} = qB \rightarrow R = \frac{m}{q} \frac{v}{B}. \quad \frac{m}{q} = \frac{1}{q/m}. \\ R = \frac{1}{5 \times 10^{7}} \frac{2 \times 10^{5}}{4/100} = \frac{1}{5 \times 10^{7}} \frac{2 \times 10^{7}}{4} = \frac{2}{5 \cdot 4} = \frac{2}{20} = \frac{1}{10} = 0.10 \, (\mathrm{m}) \\ \end{array}$$

Answer: (1) – R = 0.10 m.

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