Answer:
Talk to his team captain and ask him to alert the referee to keep a better eye on the player.
Answer:
7.8x10-12N
Explanation:
We know that
Magnetic force = F = qVB
And
Also Kinetic energy K.E is
E = (1/2)mV²
So making v subject
V = √(2E / m)
And
E = KE = 2MeV
= 2 × 106 eV
= 2 × 106 × 1.6 × 10–19 J
= 3.2 × 10–13 J
And then
V= √2x3.2E-13/1.6E-27
1.9E7m/s
Given that
mass of proton = 1.6 × 10–27 kg,
Magnetic field strength B = 2.5 T.
So F= qBv sinစ
=
So F = 1.6 × 10–19 × 2.5 × 1.9 x10^7 x sin 90°
= 7.8 x 10^-12N
As per the question the charge of one coulomb is at 0 cm of the metre stick.the second charge of 4 coulomb is situated at at 100 cm of metre stick.
hence the separation distance between them is 100 cm.
now as per the question a proton is set up between them in such a way that the net force on it is zero
let the charge of proton is q coulomb let the proton is situated at distance of x cm from the charge 1 coulomb.hence it is situated at a distance of 100-x cm from the charge 4 coulomb.
the force exerted by 1 coulomb on proton is-
the force exerted by 4 coulomb on proton is-![\frac{1}{4\pi\epsilon} \frac{q*4}{[100-x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7Bq%2A4%7D%7B%5B100-x%5D%5E2%7D)
as the net force is zero,hence-

![=\frac{1}{4\pi\epsilon} \frac{4*q}{[100-x]^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7B4%2Aq%7D%7B%5B100-x%5D%5E2%7D)
![=\frac{1}{x^2} =\frac{4}{[100-x]^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7Bx%5E2%7D%20%3D%5Cfrac%7B4%7D%7B%5B100-x%5D%5E2%7D)
![x^2=\frac{[100-x[^2}{4}[/tex[tex]x=\frac{100-x}{2}](https://tex.z-dn.net/?f=x%5E2%3D%5Cfrac%7B%5B100-x%5B%5E2%7D%7B4%7D%5B%2Ftex%3C%2Fp%3E%3Cp%3E%5Btex%5Dx%3D%5Cfrac%7B100-x%7D%7B2%7D)


cm [ans]