Question

An electron of mass 0.90×10 power -30 kg under the action of a magnetic force moves in a circle of 2.0 cm radius at a speed of 3.0×10 power 6m/sec . if proton of mass 1.8×10 power -27 kg were to move in circle of the same radius and if it were acted upon by the same magnetic field then its speed will be (1)3.0×10 power 6 m/sec (2) 1.5× 10 power 3m/sec (3) 6×10 power 4m/sec (4) can not be estimated from given data

Answer 1

Question #74684, Physics / Electromagnetism

An electron of mass $0.90 \times 10$ power $-30\,\mathrm{kg}$ under the action of a magnetic force moves in a circle of $2.0\,\mathrm{cm}$ radius at a speed of $3.0 \times 10$ power $6\,\mathrm{m/sec}$. If proton of mass $1.8 \times 10$ power $-27\,\mathrm{kg}$ were to move in circle of the same radius and if it were acted upon by the same magnetic field then its speed will be (1) $3.0 \times 10$ power $6\,\mathrm{m/sec}$ (2) $1.5 \times 10$ power $3\,\mathrm{m/sec}$ (3) $6 \times 10$ power $4\,\mathrm{m/sec}$ (4) can not be estimated from given data.

Need to find $v_p$-?

$$m_e = 9.0 \times 10^{-31} \,\mathrm{kg}$$ $$m_p = 1.8 \times 10^{-27} \,\mathrm{kg}$$ $$v_e = 3.0 \times 10^6 \,\mathrm{m/s}$$ $$|q_e| = |q_p| = 1.6 \times 10^{-19} \,\mathrm{C}$$ $$R = 2.0\,\mathrm{cm} = 2.0 \times 10^{-2} \,\mathrm{m}$$

Solution:

Lorentz force $F_L = e v B$, from picture can be seen, that $F_L = F_c$.

$$m \frac{v^2}{r} = e v B \rightarrow B = \frac{m_e v_e}{q_e R}. \text{ Let's calculate } B: B = \frac{9.0 \times 10^{-31} \cdot 3.0 \times 10^6}{1.6 \times 10^{-19} \cdot 2.0 \times 10^{-2}} = 8.4 \times 10^{-4} \,\mathrm{T}.$$

For proton $-v_p = \frac{q_p}{m_p} B R = \frac{1.6 \times 10^{-19}}{1.8 \times 10^{-27}} \cdot 8.4 \times 10^{-4} \cdot 2.0 \times 10^{-2} = 1.5 \times 10^3$.

The sign of the charge of particles, affects only the direction of force of Lorentz.

Answer: (2) $-v_p = 1.5 \times 10^3 \,\mathrm{m/s}$.

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