At a particular instant, a proton is traveling horizontally with a velocity of 531 km/s directed to the east through a region of uniform magnetic field that is directed vertically upward and has a magnitude of 21.5 mT. What is the radius of the circular path along which the proton is traveling

Explanation:

Given a proton, a proton is a positively charge particle,

q=1.609×10^-19C

The mass of proton is given as

M=1.67×10^-27kg

Velocity of the proton is

V=531km/s

Since 1km=1000m

Then, V=531,000m/s

From the Cardinal point eastward is negative direction of x

The proton is then directed eastward

Then, v=-531,000 i m/s

Magnetic field is given as

B=21.5mT

B=21.5×10^-3T

Magnetic field is directed upward i.e positive direction of y axis

Then,

B=21.5 j mT

This shows that the angle between the magnetic field and the velocity is θ=90°. They are perpendicular to each other.

Radius=?

The magnetic force is given as

F=q V×B

Note that this is a cross product and it is defined as

A×B=|A||B|Sinθ

F=qVB•Sin θ

Since the Velocity has a magnitude of

V=531,000m/s

Also B=21.5×10^-3T

Therefore,

F=qVB•Sin θ

F=1.609×10^-19×531,000×21.5×10^-3Sin90

F=1.84×10^-15N

Then, the Force to keep the body in circular motion is centripetal force

Which is given as

Fc=Mv²/R

Fc=1.67×10^-27×531,000²/R

Fc=4.71×10^-16/R

Here, the magnetic force supplies the centripetal force.

This shows that Fc=F

Fc=F

4.71×10^-16 / R = 1.84×10^-15

Cross multiply

4.71×10^-16= 1.84×10^-15 R

R=4.71^-16/1.84×10^-15

R=0.25597m

R=0.256m

Or

R=255.98mm

The radius of the circular path is approximately 0.256m

6 cos 34, 6 sin 34 trig

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