A proton moves at a speed of 2.9×107m/sat right angles to a magnetic field with amagnitude of 0.17 T. Find the magnitude of the acceleration ofthe proton. The elemental charge is 1.60×10−19C. Answer in units of m/s2.

The acceleration of the proton is .

Explanation:

Given that,

Speed of a proton,

The proton moves with this speed at right angles to a magnetic field with a magnitude of 0.17 T, B = 0.17 T

Charge on the proton,

The proton moves in the magnetic field such that the the magnetic force is balanced by the force due to its motion as :

So, the acceleration of the proton is . Hence, this is the required solution.

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

f=ma newton

f=100x2.5=250n