Question

A proton travels at right angles through a magnetic field of 0.025 Tesla's. If the magnitude of the magnetic force on the proton is 1.8x10^-14 Newton's, what is the velocity of the proton

Answer 1

Answer:

Velocity of proton will be equal to 4500 m/sec

Explanation:

We have given that proton moves right angles through a magnetic field

So angle between magnetic field and velocity of proton is $\Theta =90^{\circ}$

Magnetic field B = 0.025 Tesla

Charge on proton $e=1.6\times 10^{-16}C$

Magnetic force on the proton is given $F=1.8\times 10^{-14}N$

Magnetic force on a moving charge particle is equal to $F=qvBsin\Theta$

So $1.8\times 10^{-14}=1.6\times 10^{-16}\times v\times 0.025\times sin90^{\circ}$

v = 4500 m/sec

So velocity of proton will be equal to 4500 m/sec