An electron of mass 9.1 \times 10^{-31}\text{ kg} and charge 1.6 \times 10^{-19}\text{ C} … — JAMB UTME

An electron of mass $9.1 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$ enters perpendicularly into a uniform magnetic field of flux density $2.0 \times 10^{-3}\text{ T}$ with a speed of $3.2 \times 10^6\text{ m/s}$ . What is the radius of the circular path followed by the electron, expressed in millimeters (mm)?

Answer: 9.1 mm

Answer

The radius of the circular path followed by the electron is 9.1 mm.

When a charge

q

enters a magnetic field

B

perpendicularly at speed

v

, the magnetic force

qvB

supplies the centripetal force

\frac{mv^2}{r}

. Rearranging for radius yields

r = \frac{mv}{qB}

. Substituting

m = 9.1 \times 10^{-31}\text{ kg}

v = 3.2 \times 10^6\text{ m/s}

q = 1.6 \times 10^{-19}\text{ C}

, and

B = 2.0 \times 10^{-3}\text{ T}

gives

r = 9.1 \times 10^{-3}\text{ m} = 9.1\text{ mm}

Step-by-Step Solution

Equate the magnetic force to the centripetal force for circular motion

qvB = \frac{mv^2}{r}

A charged particle moving perpendicularly to a magnetic field experiences a magnetic force that acts entirely as a centripetal force.

Rearrange the equation to make the orbital radius

r

the subject

r = \frac{mv}{qB}

Cancelling one factor of velocity

v

from both sides allows direct computation of

r

Substitute the physical values into the formula

r = \frac{(9.1 \times 10^{-31})(3.2 \times 10^6)}{(1.6 \times 10^{-19})(2.0 \times 10^{-3})} = 9.1 \times 10^{-3}\text{ m}

Calculates the radius in standard SI units (meters).

Convert the resulting radius from meters to millimeters

r = 9.1 \times 10^{-3}\text{ m} \times 1000\text{ mm/m} = 9.1\text{ mm}

The question explicitly requests the answer in millimeters.

Key Concept

Motion of a charged particle in a uniform magnetic field

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