Question

A cosmic ray proton moving toward the Earth at 5.00 10 m/s 7 × experiences a magnetic force of 1.70 10 N −16 × . What is the strength of the magnetic field if there is a 45° angle between it and the proton’s velocity?

Answer 1

Magnetic field will be equal to
`B=3* 10^(-5)T`

Step-by-step explanation:

We have given velocity of proton
`5* 10^7m/sec`

Magnetic force experienced by proton
`F=1.7* 10^(-16)N`

Charge on proton
`q=1.6* 10^(-19)C`

Angle between field and velocity
`\Theta =45^(\circ)`

Force in magnetic field is equal to
`F=qBVsin\Theta`

So
`1.7* 10^(-16)=1.6* 10^(-19)* 5* 10^7* B* sin45^(\circ)`

`B=3* 10^(-5)T`

So magnetic field will be equal to
`B=3* 10^(-5)T`