Question

An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the electron bends along a path with a radius of curvature of 26.0 cm. Find the magnitude of the magnetic field.

Please show work. Thanks!

Answer 1

The magnitude of the magnetic field is 2.3 × 10⁻³ T.

Step-by-step explanation:

To find the magnitude of the magnetic field, we can use the formula for the radius of curvature of a charged particle in a magnetic field:

r = \frac{mv}

Where:

r is the radius of curvature

m is the mass of the particle

v is the velocity of the particle

|q| is the absolute value of the charge of the particle

B is the magnetic field strength

Plugging in the given values:

r = 26.0 cm = 0.26 m

v = \√{2 \times 1.6 \times 10⁻¹⁹ \times \frac{3000}{9.11 \times 10⁻³¹}}

Solving for the velocity:

v = 5.7 \times 10⁶ m/s

Plugging in the values for mass, charge and radius of curvature:

0.26 = \frac{5.7 \times 10⁶}{1.6 \times 10⁻¹⁹} \times B

Simplifying:

B = \frac{5.7 \times 10⁶}{1.6 \times 10⁻¹⁹} \times 0.26

B = 2.3 \times 10⁻³ T

Answer 2

The election has energy of 3000 eV entering the magnetic field.

converting to J, energy = qeV =1.6E-19*3000 where q=1.6E-19

Kinetic energy = 1/2mv^2 with m is mass of electron

1/2mv^2 = qeV

v = sqrt[1.6E-19*3000*2/9.1E-31] = 3.25E6

Magnetic force = eVB where B is magnetic field strength

Force = ma where a=radial acceleration=v^2/r

evB = mv^2/r

B = 2*3000/0.26/3.25E6 = 7.1E-3T = 71Gauss