An electron moves along the z-axis with vz=4.8×107m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (x, y, z) positions?

(2 cm, 0 cm, 0 cm)

(0 cm, 0 cm ,1 cm)

(0 cm, 2 cm , 1 cm )

Community Answer

Given: v_(z)=4.8 xx10^(7)m//s and, position of electron is at origin.(a) Here, position is: (2cm,0cm,0cm)The formula for maghetic field due to a charge particle is: B=(mu_(0)g)/(4pi)(( bar(y))xx( bar(gamma)))/(r^(2))Here, velozity is in z-direction and electron is in x-direction.{:[:. bar(v)xx bar(gamma)=vr sin(90)=v*r],[:.B=(4.9)/(4pi)(v)/(v^(2))=((4pi xx10^(-7))(-1.6 xx10^(-19)))/(4pi)([(4.8 xx10^(7))( hat(k))xx(2( hat(i)))])/((2xx10^(-2))^(2))],[:.B=(-1.6 xx ... See the full answer