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3 years ago

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A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the pro

ton and the direction of the magnetic field is 60°, what is the magnitude of the magnetic force on the proton? (e = 1.60 × 10-19
c.

2 answers:

svlad2 [7]3 years ago

8 0

Answer:

Magnitude of the force on proton = F = 1.1085 × 10^-15 N

Explanation:

Charge on proton = q = 1.60 × 10^-19 C

Velocity of proton = V = 4.0 × 10^4 m/s

Magnetic field = B = 0.20 T

Angle between V and B = θ = 60

We know that,

F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)

F = 1.1085 × 10^-15 N

Lelechka [254]3 years ago

4 0

The magnitude of the magnetic force on the proton is 1.1 × 10⁻¹⁵ N

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>magnetic force on moving charge</em><em> as follows:</em>

$\boxed{F = B q v \sin \theta}$

<em>where:</em>

<em>F = magnetic force ( N )</em>

<em>B = magnetic field strength ( T )</em>

<em>q = charge of object ( C )</em>

<em>v = speed of object ( m/s )</em>

<em>θ = angle between velocity and direction of the magnetic field </em>

Let's tackle the problem now !

$\texttt{ }$

<u>Given:</u>

speed of proton = v = 4.0 × 10⁴ m/s

magnetic field strength = B = 0.20 T

charge of proton = q = 1.60 × 10⁻¹⁹ C

direction of speed = θ = 60°

<u>Asked:</u>

the magnitude of the magnetic force on the proton = F = ?

<u>Solution:</u>

$F = B q v \sin \theta$

$F = 0.20 \times 1.60 \times 10^{-19} \times 4.0 \times 10^4 \times \sin 60^o$

$F = 6.4\sqrt{3} \times 10^{-16} \texttt{ N}$

$\boxed{F \approx 1.1 \times 10^{-15} \texttt{ N}}$

$\texttt{ }$

<h3>Learn more</h3>

Temporary and Permanent Magnet : brainly.com/question/9966993
The three resistors : brainly.com/question/9503202
A series circuit : brainly.com/question/1518810
Compare and contrast a series and parallel circuit : brainly.com/question/539204

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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