A proton moves perpendicular to a uniform magnetic field B at 1.20 107 m/s and experiences an acceleration of 2.00 1013 m/s2 in the +x direction when its velocity is in the +z direction. Determine the magnitude and direction of the field.
ma=qvBsinα
sinα=1
q=e
ma=evB
B=ma/ev =
=1.67•10⁻²⁷•2 •10¹³/1.6•10⁻¹⁹•1.2•10⁷=
=0.0173 T
F=q[v,B] =>
B is in –y direction
Well, you've stumbled upon a classic case of proton-trolling by a magnetic field! Let's see if we can make some sense out of it.
Since the proton is moving perpendicular to the magnetic field, we can use the equation F = qvB, where F is the force exerted on the proton, q is its charge, v is its velocity, and B is the field strength.
In this case, the acceleration experienced by the proton is given as 2.00 x 10^13 m/s^2, which is directed in the +x direction. The velocity of the proton is in the +z direction.
Now, because we know the acceleration is the result of the magnetic force, we can say that F = ma.
Substituting F = qvB and a = 2.00 x 10^13 m/s^2, we get qvB = ma. Rearranging this equation, we find that B = ma / (qv).
The magnitude of the field can be found by substituting the given values: B = (1.67 x 10^-27 kg)(2.00 x 10^13 m/s^2) / ((1.60 x 10^-19 C)(1.20 x 10^7 m/s)). After doing the math, we find B to be around 0.111 T.
Now, let's determine the direction of the field. Since the proton experiences an acceleration in the +x direction, it means that the magnetic force must be acting in the opposite direction, i.e., in the -x direction. Therefore, the direction of the magnetic field is in the -x direction.
So, the magnitude of the field is approximately 0.111 Tesla, and it's directed in the -x direction. Remember, though, that protons have a good sense of humor, so don't take this too seriously!
Given:
Velocity of proton, v = 1.20 x 10^7 m/s
Acceleration of proton, a = 2.00 x 10^13 m/s^2
We know that the force experienced by a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
Where:
F = force experienced by the particle (in this case, the force causing the acceleration)
q = charge of the particle (in this case, the charge of a proton, which is 1.6 x 10^-19 Coulombs)
v = velocity of the particle
B = magnetic field strength
θ = angle between the velocity vector and the magnetic field vector
In this problem, the proton experiences an acceleration in the +x direction when its velocity is in the +z direction. Since the acceleration and force are in the same direction, we can write:
F = ma
Using the equation F = qvBsinθ, we can rearrange it to solve for B:
B = F / (qvsinθ)
Since the acceleration is in the +x direction and the velocity is in the +z direction, the angle θ between them is 90° (perpendicular).
sinθ = sin(90°) = 1
Substituting the given values into the equation, we have:
B = (ma) / (qv)
B = (2.00 x 10^13 m/s^2) / ((1.6 x 10^-19 C) * (1.20 x 10^7 m/s))
B ≈ 1.04 T
Therefore, the magnitude of the magnetic field is approximately 1.04 Tesla. The direction of the magnetic field can be determined using the right-hand rule, which states that if the thumb of your right hand points in the direction of the velocity vector, and the fingers curl in the direction of the magnetic field, then the direction of the force is given by the direction in which the palm of your hand faces. Since the proton experiences a force in the +x direction, we can conclude that the magnetic field is in the +y direction.
To determine the magnitude and direction of the magnetic field, we can use the equation for the magnetic force on a charged particle moving through a magnetic field:
F = qvBsinθ
where:
- F is the magnetic force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic field strength
- θ is the angle between the velocity vector and the magnetic field vector
In this case, we are given the acceleration of the proton, which is related to the force on the proton by Newton's second law:
F = ma
where:
- m is the mass of the proton
- a is the acceleration
Since the magnetic force is the only force acting on the proton in this context, we can equate the equations for the magnetic force and acceleration:
qvBsinθ = ma
Given that the velocity is in the +z direction and the acceleration is in the +x direction, we can conclude that the angle θ between the velocity and the magnetic field vectors is 90 degrees (perpendicular).
Plugging in the given values:
q(1.6 x 10^-19 C)(1.2 x 10^7 m/s)Bsin90° = (1.67 x 10^-27 kg)(2.0 x 10^13 m/s^2)
Simplifying:
(1.92 x 10^-12 B) = (3.34 x 10^-14 kg)(m/s^2)
Now we can solve for B:
B = [(3.34 x 10^-14 kg)(m/s^2)] / (1.92 x 10^-12)
B ≈ 1.74 T
So, the magnitude of the magnetic field is approximately 1.74 T.
Since the proton experiences an acceleration in the +x direction, we can determine the direction of the magnetic field using the right-hand rule. By convention, the direction of the magnetic field is directed away from your palm if you curl your fingers in the direction of motion and the thumb points in the direction of the acceleration.
Therefore, the direction of the magnetic field is in the -y direction.