An alpha particle carrying a positive charge of +3.2 \times 10^{-19}\text{ C} moves horizo… — JAMB UTME

An alpha particle carrying a positive charge of $+3.2 \times 10^{-19}\text{ C}$ moves horizontally along the $+x$ -axis at a velocity of $5.0 \times 10^6\text{ m/s}$ . It enters a velocity selector region containing a uniform electric field of $4.0 \times 10^4\text{ V/m}$ directed along the $+y$ -axis. What magnitude and direction of uniform magnetic field $\vec{B}$ are required for the particle to traverse the region undeflected?

$8.0 \times 10^{-3}\text{ T}$ directed along the $+z$ -axis (out of the page)Answer
B
$8.0 \times 10^{-3}\text{ T}$ directed along the $-z$ -axis (into the page)
C
$2.0 \times 10^{11}\text{ T}$ directed along the $+z$ -axis (out of the page)
D
$1.25 \times 10^{-2}\text{ T}$ directed along the $-y$ -axis

Answer

The required magnetic field has a magnitude of

8.0 \times 10^{-3}\text{ T}

directed along the

+z

-axis (out of the page).

The condition for undeflected movement through crossed electric and magnetic fields (velocity selector) requires the magnetic force to be equal and opposite to the electric force. The magnitude is

B = E / v = (4.0 \times 10^4) / (5.0 \times 10^6) = 8.0 \times 10^{-3}\text{ T}

. For a positively charged particle moving along the

+x

-axis with an electric force in the

+y

-direction, Fleming's left-hand rule (or right-hand vector cross product) specifies that the magnetic field must point out of the page along the

+z

-axis to direct the magnetic force along the

-y

-direction.

Step-by-Step Solution

Set up the force balance condition for undeflected motion.

Net force

F_{net} = F_E - F_B = 0 \implies qE = qvB \sin\theta

. Since

\vec{v}

and

\vec{B}

are perpendicular,

\sin\theta = 1

, giving

qE = qvB

For zero deflection, the electric force and magnetic force must be equal in magnitude and opposite in direction.

Calculate the magnitude of the magnetic field

B

B = \frac{E}{v} = \frac{4.0 \times 10^4\text{ V/m}}{5.0 \times 10^6\text{ m/s}} = 8.0 \times 10^{-3}\text{ T}

Dividing electric field strength by particle speed yields the required magnetic field strength.

Determine the direction of the magnetic field using vector cross-product rules.

Electric force

\vec{F}_E = q\vec{E}

points in the

+y

-direction. Thus, magnetic force

\vec{F}_B = q(\vec{v} \times \vec{B})

must point in the

-y

-direction. With

\vec{v}

in the

+x

-direction (

\hat{i}

\hat{i} \times \hat{k} = -\hat{j}

, so

\vec{B}

must point along the

+z

-axis (out of the page).

Opposing vector directions are necessary to achieve equilibrium between electric and magnetic forces.

Key Concept

Velocity Selector and Lorentz Force Equilibrium

Estimated Time:2m 0s

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