Question

An alpha-particle (charge $2e$) moves along a circular path of radius $1\stackrel{0}{A}$ with a uniform speed of $2\times {10}^{6}m{s}^{-1}$. Calculate the magnetic field produced at the centre of the circular path

Answer 1

$Givenspeedofalphaparticle=2\times {10}^{6}m/sConstantvalueofunitcharge\left(e\right)=1.6\times {10}^{-19}CRadius=1\dot{A}=1\times {10}^{-10}mSo2e=3.2\times {10}^{-19}Coulomb.AsmagneticfieldatcentreofcirclarloopB=\frac{{\mu }_{0}I}{2R};where{\mu }_0=4\pi \times {10}^{-7}Tm/AAlsoCurrentI=\frac{2e}{T};T=Timeperiod=\frac{2\pi R}{V}=\frac{2\pi \times 1\times {10}^{-10}}{2\times {10}^6}=3.14\times {10}^{-16}sI=\frac{3.2\times {10}^{-19}}{3.14\times {10}^{-16}}=1.01\times {10}^{-3}AThereforeB=\frac{4\pi \times {10}^{-7}\times 1.01\times {10}^{-3}}{2\times {10}^{-10}}=6.34$