Answer :
Answer:[tex]0.33 m/s^2[/tex]
Explanation:
Given
mass of charge[tex]=1.81\times 10^{-3} kg[/tex]
charge on Particle(q)[tex]=1.22\times 10^{-8} C[/tex]
velocity [tex]\vec{v}=3\times 10^4\hat{j}[/tex]
Magnetic Field[tex]=1.63\hat{i}+0.98\hat{j}[/tex]
Force on a charge Particle in a Magnetic Field
[tex]F=q\left ( \vec{v}\times \vec{B}\right )[/tex]
[tex]a=\frac{F}{m}=\frac{q\left ( \vec{v}\times \vec{B}\right )}{m}[/tex]
[tex]a=\frac{q\left ( \left ( 3\times 10^4\hat{j}\right )\times \left ( 1.63\hat{i}+0.98\hat{j}\right )\right )}{1.81\times 10^{-3}}[/tex]
[tex]a=\frac{1.22\times 10^{-8}\times 4.89\times 10^4\hat{k}}{1.81\times 10^{-3}}[/tex]
[tex]a=0.33\hat{k} m/s^2[/tex]
So acceleration is directed towards - z direction considering [tex]\hat{i}[/tex] as +x axis and [tex]\hat{j}[/tex] as + y axis
The magnitude and direction of the particle’s acceleration is about 0.33 m/s² in negative z-axis.
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Further explanation
Let's recall magnetic force on moving charge as follows:
[tex]F = B q v \sin \theta[/tex]
where:
F = magnetic force ( N )
B = magnetic field strength ( T )
q = charge of object ( C )
v = speed of object ( m/s )
θ = angle between velocity and direction of the magnetic field
Let's tackle the problem now !
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Given:
speed of particle = v_y = 3.00 × 10⁴ m/s
magnetic field strength = B_x = 1.63 T
charge of particle = q = 1.22 × 10⁻⁸ C
direction of speed = θ = 90°
mass of particle = m = 1.81 × 10⁻³ kg
Asked:
the magnitude and direction of the particle's acceleration = a = ?
Solution:
[tex]a = F \div m[/tex]
[tex]a = ( B_x q v_y \sin \theta ) \div m[/tex]
[tex]a = (1.63 \times 1.22 \times 10^{-8} \times 3.00 \times 10^4 \times \sin 90^o) \div ( 1.81 \times 10^{-3} )[/tex]
[tex]\boxed{a \approx 0.33 \texttt{ m/s}^2}[/tex]
According to the Left Hand Rule , the direction of acceleration is in negative z-axis.
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Learn more
- Temporary and Permanent Magnet : https://brainly.com/question/9966993
- The three resistors : https://brainly.com/question/9503202
- A series circuit : https://brainly.com/question/1518810
- Compare and contrast a series and parallel circuit : https://brainly.com/question/539204
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Answer details
Grade: High School
Subject: Physics
Chapter: Magnetic Field
