In a cathode-ray experiment, a beam of electrons passes undeflected through mutually perpe… — JAMB UTME

In a cathode-ray experiment, a beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field intensity is $4.0 \times 10^4\text{ V m}^{-1}$ and the magnetic flux density is $2.0 \times 10^{-3}\text{ T}$ , what is the speed of the electrons in the beam?

$2.0 \times 10^7\text{ m s}^{-1}$ Answer
B
$8.0 \times 10^1\text{ m s}^{-1}$
C
$5.0 \times 10^{-8}\text{ m s}^{-1}$
D
$2.0 \times 10^1\text{ m s}^{-1}$

Answer

The speed of the electrons in the beam is

2.0 \times 10^7\text{ m s}^{-1}

When a cathode ray beam passes undeflected through perpendicular electric (

E

) and magnetic (

B

) fields, the electric force (

eE

) and magnetic force (

evB

) are equal in magnitude and opposite in direction. Equating

eE = evB

simplifies to

v = \frac{E}{B}

. Substituting

E = 4.0 \times 10^4\text{ V m}^{-1}

and

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

yields an electron speed of

2.0 \times 10^7\text{ m s}^{-1}

Step-by-Step Solution

Identify the condition for no deflection in crossed electric and magnetic fields.

The electrostatic force

F_e = eE

equals the magnetic force

F_b = evB

For zero net deflection, the opposing electric and magnetic forces must balance each other exactly.

Rearrange the force balance equation to solve for speed

v

v = \frac{E}{B}

Canceling the charge

e

on both sides of

eE = evB

isolates velocity

v

as the ratio of electric field to magnetic field.

Substitute the given values into the velocity equation.

v = \frac{4.0 \times 10^4\text{ V m}^{-1}}{2.0 \times 10^{-3}\text{ T}} = 2.0 \times 10^7\text{ m s}^{-1}

Evaluating

\frac{4.0}{2.0} \times 10^{4 - (-3)}

gives

2.0 \times 10^7\text{ m s}^{-1}

Key Concept

Velocity selector condition for cathode rays in crossed electric and magnetic fields

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