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

Answer:

Magnetic field will be equal to $B=3\times 10^{-5}T$

Explanation:

We have given velocity of proton $5\times 10^7m/sec$

Magnetic force experienced by proton $F=1.7\times 10^{-16}N$

Charge on proton $q=1.6\times 10^{-19}C$

Angle between field and velocity $\Theta =45^{\circ}$

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

So $1.7\times 10^{-16}=1.6\times 10^{-19}\times 5\times 10^7\times B\times sin45^{\circ}$

$B=3\times 10^{-5}T$

So magnetic field will be equal to $B=3\times 10^{-5}T$