Scientists study a radioactive source that produces a stream of positively-charged particles moving with different speeds along
2 answers:
Answer:
A)
Explanation:
A) the particles do not reflect the electric and magnetic forces, they must have the same magnitude and opposite direction.
the magnetic force is perpendicular to the speed that is on the x axis and to the magnetic field that stakes the z axis, so the force goes in the direction of the positive y axis.
With this force it must be opposite to the electric force, this implies that the electric field must go in the direction of the negative y axis.
Let's use Newton's equation of equilibrium
= 0
FM = Fe
.q v B = q E
E = v B
E = 2 0.1
E = 0.2 N / C
.b) when the particular one reaches the camera with a magnetic field, the magnetic force perpendicular to the speed causes them to start a uniform circular movement since the magnitude of the speed does not change.
With the magnetic force it is on the axis and the circle described by the particles is on this axis
.c) The particles are in a circular motion the point where their motion is reversed they move in the negative direction when they are at the lowest or highest part of the path or I know that they have traveled
Let's use kinematics
.v = va - a t
.v = -vow
-vow = vow - at
.a = 2 vow / t
.a = 2 2 / 0.1 10-3
.a = 40 103 m / s2
Now let's use Newton's second law
FM = m a
Q v B = m a
.m = q v B / a
.m = 1.6 10-19 2.0 0.1 / 40 103
.m = 8 10-25 kg
Answer:
a) The strength of the electric field is 0.2 V/m and its direction is -y axis.
b) The particle moves in a circular path because the magnetic force is perpendicular to its speed.
c) The mass of the particle is 5.09x10⁻²⁵ kg
Explanation:
a) If the magnetic field has a z direction and the speed is towards x, therefore, the magnetic force has a y direction. Because of this, so that the particle does not deviate, the electric field must be in the direction -y. The magnitude is equal to:
E = B*V = 0.1*2 = 0.2 V/m
b) The particle moves in a circular path because the magnetic force is perpendicular to its speed.
c) The time period is:
T = 2*t = 2*0.1 = 0.2 ms
The mass of the particle is:
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Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
Answer:
- The magnitude of the vector
is 107.76 m
Explanation:
To find the components of the vectors we can use:
where is the magnitude of the vector, and θ is the angle over the positive x axis.
The negative x axis is displaced 180 ° over the positive x axis, so, we can take:
Now, we can perform vector addition. Taking two vectors, the vector addition is performed:
So, for our vectors:
To find the magnitude of this vector, we can use the Pythagorean Theorem
And this is the magnitude we are looking for.