A α-particle enters in a magnetic field of strength (3î + 2ĵ) with velocity (5 × 10 ⁵ î). The Magnetic force experienced by the particle will be :

Question

A &alpha;-particle enters in a magnetic field of strength (3&icirc; + 2&jcirc;) with velocity (5 &times; 10 5 &icirc;). The Magnetic force experienced by the particle will be :

Prepp Team · Accepted Answer

Calculating Magnetic Force on an Alpha Particle

The question asks us to find the magnetic force experienced by an alpha particle moving in a given magnetic field with a specific velocity. This force is determined by the Lorentz force formula for a charged particle in a magnetic field.

Understanding the Lorentz Force

The magnetic force ($\vec{F}$) on a charged particle with charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the vector product:

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

This formula tells us that the force is perpendicular to both the velocity vector ($\vec{v}$) and the magnetic field vector ($\vec{B}$). The magnitude of the force depends on the charge, the speed, the magnetic field strength, and the angle between $\vec{v}$ and $\vec{B}$.

Given Information

Charge of an alpha particle ($q$): An alpha particle consists of two protons and two neutrons. Its charge is $+2e$, where $e$ is the elementary charge ($1.6 \times 10^{-19}$ C). So, $q = 2 \times (1.6 \times 10^{-19} \text{ C}) = 3.2 \times 10^{-19}$ C.
Velocity of the alpha particle ($\vec{v}$): $(5 \times 10^5 \hat{i})$ m/s. This means the particle is moving along the positive x-axis.
Magnetic field ($\vec{B}$): $(3 \hat{i} + 2 \hat{j})$ T. The magnetic field has components along the positive x and positive y axes.

Calculating the Cross Product $\vec{v} \times \vec{B}$

We need to calculate the cross product of the velocity and magnetic field vectors:

$$ \vec{v} \times \vec{B} = (5 \times 10^5 \hat{i}) \times (3 \hat{i} + 2 \hat{j}) $$

Using the distributive property of the cross product:

$$ \vec{v} \times \vec{B} = (5 \times 10^5 \hat{i}) \times (3 \hat{i}) + (5 \times 10^5 \hat{i}) \times (2 \hat{j}) $$

$$ \vec{v} \times \vec{B} = (5 \times 10^5 \times 3) (\hat{i} \times \hat{i}) + (5 \times 10^5 \times 2) (\hat{i} \times \hat{j}) $$

Recall the standard unit vector cross products:

$\hat{i} \times \hat{i} = 0$
$\hat{i} \times \hat{j} = \hat{k}$
$\hat{i} \times \hat{k} = -\hat{j}$

Applying these rules:

$$ \vec{v} \times \vec{B} = (15 \times 10^5) (0) + (10 \times 10^5) (\hat{k}) $$

$$ \vec{v} \times \vec{B} = 0 + 10 \times 10^5 \hat{k} = 10^6 \hat{k} \text{ (units of m/s } \cdot \text{ T)} $$

Calculating the Magnetic Force $\vec{F}$

Now, we use the Lorentz force formula $\vec{F} = q (\vec{v} \times \vec{B})$, substituting the charge $q$ and the calculated cross product:

$$ \vec{F} = (3.2 \times 10^{-19} \text{ C}) \times (10^6 \hat{k} \text{ m/s} \cdot \text{ T}) $$

$$ \vec{F} = (3.2 \times 10^{-19} \times 10^6) \hat{k} \text{ N} $$

$$ \vec{F} = 3.2 \times 10^{-13} \hat{k} \text{ N} $$

The magnetic force experienced by the alpha particle is $3.2 \times 10^{-13} \hat{k}$ Newtons.

Comparing with Options

Let's compare our calculated force with the given options:

Option 1: $3.2 \times 10^{-13} \hat{k}$
Option 2: $1.6 \times 10^{-13} \hat{k}$
Option 3: $6.4 \times 10^{-13} \hat{k}$
Option 4: $5.2 \times 10^{-13} \hat{k}$

Our calculated force matches Option 1.

Prepp Team · Answer

1.6 &times; 10 -13  k̂

Prepp Team · Answer

6.4 &times; 10 -13k̂

Prepp Team · Answer

5.2 &times; 10-13  k̂