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

For plato users: Option E.

Step-by-step explanation:

Answer 2

Velocity of proton will be equal to 4500 m/sec

Step-by-step 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* 10^(-16)C`

Magnetic force on the proton is given
`F=1.8* 10^(-14)N`

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

So
`1.8* 10^(-14)=1.6* 10^(-16)* v* 0.025* sin90^(\circ)`

v = 4500 m/sec

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