Ask question

Ask question

All categories

3 years ago

10

The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the

field with a velocity ~v = vx ˆı+vy ˆ+vz ˆk, where vx = 2 m/s, vy = 6 m/s and vz = 8 m/s. The charge on the electron is −1.602 × 10−19 C. What is the ˆı component of the force exerted on the electron by the magnetic field? Answer in units of N.

1 answer:

krek1111 [17]3 years ago

8 0

Answer:

Explanation:

The force exerted in a magnetic field is given as

F = q (v × B)

Where

F is the force entered

q is the charge

v is the velocity

B is the magnetic field

Given that,

The magnetic field is

B = 2•i + 4•j. T

The velocity of the electron is

v = 2•i + 6•j + 8•k. m/s

Also, the charge of an electron is

q = -1.602 × 10^-19 C.

Then note that,

V×B is the cross product of the speed and the magnetic field

Then,

F = q (V×B)

F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)

Note

i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]

F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]

F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)

F = -1.602 × 10^-19(16•j - 32•i)

F = -1.602 × 10^-19 × ( -32•i + 16•j)

F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j

Then, the x component of the force is

Fx = 5.126 × 10^-18 N

Also, the y component of the force is

Fy = -2.563 × 10^-18 N

You might be interested in

The rate of chage of displacement is called __________

faltersainse [42]

Answer:

Velocity is the rate of change of displacement.

8 0

2 years ago

Read 2 more answers

How does a sound wave's amplitude affect the sound we hear? How does a sound

damaskus [11]

Answer:

cause we hear differently

Explanation:

7 0

3 years ago

Read 2 more answers

To calculate the velocity of an object, the of the position vs time graph should be calculated.

Firdavs [7]

The slope of the graph

3 0

3 years ago

How do you solve this problem?

Katarina [22]

The particle has acceleration vector

$\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath$

We're told that it starts off at the origin, so that its position vector at $t=0$ is

$\vec r_0=\vec0$

and that it has an initial velocity of 12 m/s in the positive $x$ direction, or equivalently its initial velocity vector is

$\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath$

To find the velocity vector for the particle at time $t$ , we integrate the acceleration vector:

$\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau$

$\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath$

$\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath$

Then we integrate this to find the position vector at time $t$ :

$\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau$

$\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath$

$\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

Solve for the time when the $y$ coordinate is 18 m:

$18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s$

At this point, the $x$ coordinate is

$\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m$

so the answer is C.

7 0

3 years ago

a ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. what is the balls overall s

padilas [110]

Answer:

30m/ s

Explanation:

this is an example of a projectile, where the path traced by the ball will be a downward parabola.

one must know that the <em>vertical velocity of an object having projectile motion is zero at the top of its trajectory</em> and that the <em>horizontal velocity remains constant throughout</em>.

this means, at the top of the trajectory the <u>ball's overall speed will be equal to its horizontal velocity</u>.

and horizontal velocity is given 30 m/s

hence, overall speed of the ball = 30m/ s

8 0

3 years ago