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Nutka1998 [239]
2 years ago
15

An electron moving with a speed of 4.0 * 105 m>s in the positive x direction experiences zero magnetic force. When it moves i

n the positive y direction, it experiences a force of 3.2 * 10-13 N that points in the positive z direction. What are the direction and magnitude of the magnetic field
Physics
2 answers:
neonofarm [45]2 years ago
7 0

Answer:

B= 4.99T in negative x direction

Explanation:

F= qvBsin∅

q: charge on particle= 1.602 × 10^-19

v: velocity of the particle= 4 × 10⁵

B: Magentic field=???

∅: angle between magnetic field and direction of motion of charge particle

When moving in positive x- direction, the force is zero. According to the equation, the force will be zero when angle between direction of motion and magnetic field is either zero or 180⁰.

Now, when moving in positive y-direction, the force is 3.2 ˣ 10^-13 in positive z direction. and ∅=90. From Flemming's left hand rule, we can deduce that magnetic field is in negative x direction

3.2 ˣ 10^-13= 1.602 × 10^-19 × 4 × 10⁵ × B × sin90

B= 4.99T

IgorLugansk [536]2 years ago
4 0

Answer:

Explanation:

Given that,

Charge on an electron is

q=1.609×10^-19C

Speed of electron V=4×10^5 m/s in positive x direction

V = 4×10^5 •i m/s

At this point F=0N

When the electron is moving in positive y direction

V =4×10^5 •j m/s

F = 3.2×10^-13 N in positive z direction.

F = 3.2×10 ^-13 •k N

What is the direction and magnitude of the magnetic field B?

The force in a magnetic field is given as

F = q (V ×B)

The only time V×B is zero is when both V and B are in same direction

This shows that B is in the positive x direction or negative x direction

i.e B= Bx •i

Now applying this to the second condition

F= q(V×B)

3.2×10^-13 •k =1.609×10^-19 × (4×10^5•j ×Bx •i)

Note that, j×i =-k

3.2×10^-13 •k / 1.609×10^-19 = (4×10^5Bx •(j×i))

1988812.927 •k =- 4×10^5Bx•k

1988812.927/4×10^5 •k =-Bx •k

4.972 •k =-Bx •k

Then, Bx = -4.972 •k Telsa

Then, the magnitude of B is

B=4.972 T

And its direction is negative z direction.

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Answer:

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The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

\large {\boxed {a_s = v^2 / R} }

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

F = ma

F = m\frac{v^2}{R}

7.80w = m\frac{v^2}{R}

7.80mg = m\frac{v^2}{R}

7.80g = \frac{v^2}{R}

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\omega = \frac{v}{2 \pi R}  → in rev/s

\omega = \frac{29}{2 \pi \times 11.0}

\omega \approx 0.42 ~ rev/s

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Uniform Circular Motion : brainly.com/question/2562955
  • Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

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