Question

Two protons move parallel to each other with speed V1 and va
Net force between them​

Answer 1

The force of magnetic induction is given by

Fmag=e(v×B);B=2πμ0r3e(v×r)

∴Fmag=2πμ0×r3e2[v×(v×r)]

=2πμ0×r3e2[(v.r)×v−(v.v)×r

= 1

Answer 2

Concept:-

Net force means the sum of total forces applied to the particle. In the matter of proton, there are two forces.

Given:-

speed of 1st proton $v1$ and speed of 2nd proton $va$

Find:-

We have to find the net force between them.

Solution:-

The electric force between two protons,

$F_{e} = \frac{1}{4\piε _{0} }$  ×$\frac{q^{2} }{r^{2} }$

Now, the magnetic force is,

$F_{m}= q(v1-va)B$

B is the magnetic field

B$=\frac{1}{4\pi u_{0} } \frac{IdL}{r^{2} }$

Now $F_{m}=q(v1-va)\frac{1}{4\pi u_{0} } \frac{q(v1-va)}{r^{2} }$

or,$F_{m}=\frac{1}{4\pi u_{0} } \frac{q^{2} (v1-va)^{2} }{r^{2} }$

∴$F_{net} = F_{e}+ F_{m}$

or,$F_{net} = \frac{q^{2}} {4\pi r^{2} }[\frac{1}{e_{0} }+ \frac{(v1-va)^{2} }{u_{0} } ]$

Hence, the net force is $F_{net} = \frac{q^{2}} {4\pi r^{2} }[\frac{1}{e_{0} }+ \frac{(v1-va)^{2} }{u_{0} } ]$.