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)

Answer 1

Answer:

$9.198\times 10^6 m/s$

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

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

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

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

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