## . A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C

Question

. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C and mp = 1.67 · 10-27 kg)

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1 year 2021-09-04T13:01:40+00:00 1 Answers 141 views 0

$$9.198\times 10^6 m/s$$

Explanation:

We are given that

Magnetic field, B=1.2 T

Radius of circular path, r=0.080 m

$$q_p=1.6\times 10^{-19} C$$

$$m_p=1.67\times 10^{-27} kg$$

$$\theta=90^{\circ}$$

We have to find the speed of proton.

We know that

Magnetic force, F=$$qvBsin\theta$$

According to question

Magnetic force=Centripetal force

$$q_pvBsin90^{\circ}=\frac{m_pv^2}{r}$$

$$1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}$$

$$v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}$$

$$v=9.198\times 10^6 m/s$$