Question

  A proton is moving east at 4x106 m/s. Calculate the strength of the magnetic field at a distance of 1cm north. 

Answer 1

When a proton enters a magnetic field it moves in circular path , the formula of radius on that circular path is given by -

$R=$ $\dfrac{mv}{qB}$

Mass of proton m = 1.6726$\times 10^{-27}kg$

charge of proton q = +1.6$\times 10^{-19}\ C$

velocity of proton v = 4$\times10^{6}\ m/s$

radius of path R = $10^{-2} \ m$

$10^{-2}=\dfrac{1.6726\times 10^{-27}\times4\times 10^{6}}{1.6\times 10^{-19}\times B}$

$B = \dfrac{1.6726\times 10^{-27}\times 4\times 10^{6}}{1.6\times 10^{-19}\times 10^{-2}}=$ 4.18 T