Answer to Question #159460 in Electricity and Magnetism for Saurabh

Question

A particle moves in a slowly varying magnetic field 𝐵⃗ = 𝐵𝑧(𝑥)kcap and a uniform magnetic field and in small electricfield 𝐸⃗ = 𝐸𝑦jcap. Calculate the drifts. Show that change in perpendicular kinetic energy can be accounted for by the drift along the electric field. Calculate the rate of change of (1/2)m(V perpendicular)^2

.

Answer 1

Given,

Varying magnetic field $\overrightarrow{(B)}=Bz(x)\hat{k}$

Electric field $\overrightarrow{(E)}=Ey\hat{j}$

Let the mass of the particle be m and charge on the particle be q

Let charge is entering into the electric and magnetic field along to the x direction.

So, force on the charge $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$=qv Bz(x)(\hat{-j})$

Now, electrostatic force is working along the y axis because electric field is along to the y axis

So, the electrostatic force along to they axis $\overrightarrow{F_e}=qEy\hat{j}$

but here we can say that electrostatic force and the magnetic force are opposite to each other

but it is given in the question that particle is moving with constant velocity it means electric force = magnetic force on the particle

$qEy=qvBz(x)$

From the above we can conclude that

$v=\frac{Ey}{Bz(x)}$

Kinetic energy of the particle $E_{KE}=\frac{mv^2}{2}$

Now, substituting the value of v in the above,

$E_{KE}=\frac{m(\frac{Ey}{Bz(x)})^2}{2}$

$\Rightarrow E_{KE}=\frac{m(Ey)^2}{2(Bz(x))^2}$

$\frac{dE_{KE}}{dt}=\frac{m}{2}[\frac{2(Ey)^2Bz(x)\frac{dx}{dt}-Bz(x)\frac{Edy}{dt}}{(Bz(z))^2}]$