Question

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 C)

Answer 1

Answer:

Radius = 0.11 m

Explanation:

To find the speed of the proton we know that

$KE = PE$

here we have

$\frac{1}{2}mv^2 = qV$

now we have

$v = \sqrt{\frac{2qV}{m}}$

now we have

$v = \sqrt{\frac{2(1.60 \times 10^{-19})(1000)}{(1.67\times 10^{-27})}}$

$v = 4.38 \times 10^5 m/s$

Now for the radius of the circular motion of charge we know

$\frac{mv^2}{R} = qvB$

$R = \frac{mv}{qB}$

$R = \frac{(1.67\times 10^{-27})(4.38 \times 10^5)}{(1.60\times 10^{-19})(0.040)}$

$R = 0.11 m$

Answer 2

The radius of the proton orbit is 0.114m

kinetic energy:

When the proton travels through the potential V it gains kinetic energy given below:

$\frac{1}{2}mv^2=qV\\\\v=\sqrt[]{\frac{2qV}{m} }\\\\v=\sqrt{\frac{2\times (1.6\times10^{-19})\times10^3}{1.67\times10^{-27}}$

$v=4.37\times10^5m/s$

magnetic force:

Now, a moving charge under magnetic field B undergoes circular motion due to the magnetic force being perpendicular to the velocity of the charge, given by:

$\frac{mv^2}{r}=qvB$

$r=\frac{mv}{qB} \\\\r=\frac{1.67\times10^{-27}\times4.37\times10^5}{1.6\times10^{-19}\times0.040} m\\\\r=0.114m$

the radius of the orbit is 0.114m

Learn more about magnetic force:

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