Question

a proton moving with a velocity of 2.5x10^7 m/s enters magnetic field of 2.5T making an angle 30 with magnetic field. the force on proton is?

Answer 1

hey mate here is ur solution

Given b = 2.5T, v = 2.5x10^7,theta = 30deg,q = 1.6x10^-19

therefore F = 2.5 x 1.6x10^-19 x 2.5x10^& = 5x10^-12N

Answer 2

Answer:

A proton enters a magnetic field of intensity 2.5T with a velocity of 2.5x10^7 m/s making an angle of 30 with the magnetic field, the force on the proton is $5.14*10^{-12}N$

Explanation:

• When a particle with charge $q$ moving with velocity $V$ through the magnetic field by making an angle $\alpha$ with the magnetic field, the particle experiences a force called Lorentz force, which is given by the formula

        $F=qVB sin\alpha$

From the question, we have given

velocity of proton $V=2.5*10^{7} m/s$

The intensity of the magnetic field $B=2.5T$

the charge of the proton $q=1.6*10^{-19}C$

the angle made by the proton with the magnetic field is  $\alpha =30$

after the substitution

⇒

$F=1.6*10^{-19}*2.57*10^{7}*2.5*sin30=5.14*10^{-12} N$