Question

What uniform magnetic field, applied perpendicular to a beam of electrons moving at $\mathbf{1}\mathbf{.30}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$ m/s, is required to make the electrons travel in a circular arc of radius 0.350 m?

Short answer

The required magnetic field is$2.11\times {10}^{-5}T$ .

Step-by-step solution

Listing the given quantities

• Velocity isV = .$1.3\times {10}^6\text{\hspace{0.17em}m/s}$
• Radius is r =.$0.350m$

Understanding the concept of centripetal force and magnetic field

Weuse the equilibrium condition of the magnetic force on the electron and centripetal force due to the circular path.

Formula:

$F_m=qvB$

$F_c=\frac{m{v}^2}{r}$

 Step 3: Explanation

For the circular motion of an electron:

$\begin{array}{c}{F}_m=F_c\\ qvB=\frac{m{v}^2}{r}\\ B=\frac{mv}{rq}\end{array}$

Substituting the given values in the above expression, and we get,

$\begin{array}{c}B=\frac{9.11\times {10}^{-31}\times 1.3\times {10}^6}{0.35\times 1.6\times {10}^{-19}}\\ =2.11\times {10}^{-5}T\end{array}$

The required magnetic field is$2.11\times {10}^{-5}T$ .