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Romashka-Z-Leto
17 days ago
6

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^?
Physics
2 answers:
ValentinkaMS [2.4K]17 days ago
8 0

The particle’s acceleration magnitude and direction in a uniform magnetic field B =(1.63T)i^+(0.980T)j^ is \bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}

Explanation:

A particle weighing 1.81×10−3 kg with a charge of 1.22×10−8 C has a velocity v⃗ =(3.00×104m/s)j^ at a specific moment. What are the acceleration's magnitude and direction from a uniform magnetic field B =(1.63T)i^+(0.980T)j^?

A charged particle possesses an electric charge. Electric charge is a property that results in force attraction or repulsion in an electromagnetic field. A uniform magnetic field occurs when the magnetic field lines are straight and parallel, ensuring a consistent magnetic force throughout the area.

According to Newton's second law, the force can be defined by:

F=ma

The magnetic force is given by

F= qv \times B

ma = qv \times B

a = \frac{qv \times B}{m}

By substituting the earlier values, we find:

a = \frac{(1.22 \times 10^{-8} C) (3 \times 10^{4} m/s) (1.63 T ) (\hat{j} \times \hat{i})}{1.81 \times 10^{-3} kg} = -(0.330 m/s^2) \bold{\hat{{k}}}

Thus, the particle's resulting acceleration in a uniform magnetic field B =(1.63T)i^+(0.980T)j^ is confirmed to be \bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}

Learn further about the particle's acceleration

ValentinkaMS [2.4K]17 days ago
7 0

Answer:

The particle's acceleration magnitude and direction is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given:

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

To find the particle’s acceleration

The formula for acceleration is

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We will calculate v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, substitute all known values into the formula for acceleration

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

a= -0.3296\ \hat{k}\ m/s^2

The negative sign indicates the direction is opposite.

Thus, the particle’s acceleration magnitude and direction is a= 0.3296\ \hat{k}\ m/s^2

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