Question

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 proton 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.

Answer 1

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    

Answer 2

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

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Further explanation

Let's recall magnetic force on moving charge as follows:

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

where:

F = magnetic force ( N )

B = magnetic field strength ( T )

q = charge of object ( C )

v = speed of object ( m/s )

θ = angle between velocity and direction of the magnetic field

Let's tackle the problem now !

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Given:

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°

Asked:

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

Solution:

$F = B q v \sin \theta$

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

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

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

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field