Question

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 1

The magnitude and direction of the particle’s acceleration produced by 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 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 $B =(1.63T)i$^$+(0.980T)j$^?

A charged particle is a particle with an electric charge. Whereas electric charge is the matter physical property that causes to experience a force when placed in an electromagnetic field. Uniform magnetic field is the condition when magnetic field lines are parallel then magnetic force experienced by an object is same at all points in that field

From Newton's second law, the force is given by:

$F=ma$

Magnetic force is

$F= qv \times B$

$ma = qv \times B$

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

Subsituting with the givens above we get

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

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

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Answer 2

Answer:

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

Explanation:

Given that,

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

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

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

We need to calculate the value of $v\times B$

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

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

Now, put the all values into the acceleration 's formula

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

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

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$