contestada

What uniform magnetic field, applied perpendicular to a beam
ofelectrons moving at 1.30 x 106 m/s, is required to
makethe elctrons travel in a ciruclar arc of radius 0.350 m?

Respuesta :

Answer:

The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]

Explanation:

Given that,

Speed [tex]v=1.30\times10^{6}\ m/s[/tex]

Radius = 0.350 m

We need to calculate the magnetic field

Using formula of magnetic field

[tex]B =\dfrac{mv}{qr}[/tex]

Where, m = mass of electron

v = speed of electron

q = charge of electron

r = radius

Put the value into the formula

[tex]B=\dfrac{9.1\times10^{-31}\times1.30\times10^{6}}{1.6\times10^{-19}\times0.350}[/tex]

[tex]B=2.11\times10^{-5}\ T[/tex]

Hence, The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]