To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^?

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

johanrusli

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 !

[tex]\texttt{ }[/tex]

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

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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