The magnetic force experienced by the proton is given by
![F=qvB \sin \theta](https://tex.z-dn.net/?f=F%3DqvB%20%5Csin%20%5Ctheta)
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so
![\sin \theta=1](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D1)
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
So we have
![ma=qvB](https://tex.z-dn.net/?f=ma%3DqvB)
from which we can find the magnitude of the field:
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved mirror than the centre of curvature.</em>
<em></em>
During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Answer:
Magnitude of vector A = 0.904
Explanation:
Vector A , which is directed along an x axis, that is
![\vec{A}=x_A\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3Dx_A%5Chat%7Bi%7D)
Vector B , which has a magnitude of 5.5 m
![\vec{B}=x_B\hat{i}+y_B\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BB%7D%3Dx_B%5Chat%7Bi%7D%2By_B%5Chat%7Bj%7D)
![\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25](https://tex.z-dn.net/?f=%5Csqrt%7Bx_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%7D%3D5.5%5C%5C%5C%5Cx_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%3D30.25)
The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A ![\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%2B%5Cvec%7BB%7D%3D6x_A%5Chat%7Bj%7D%5C%5C%5C%5Cx_A%5Chat%7Bi%7D%2Bx_B%5Chat%7Bi%7D%2By_B%5Chat%7Bj%7D%3D6x_A%5Chat%7Bj%7D)
Comparing we will get
![x_A=-x_B\\\\y_B=6x_A](https://tex.z-dn.net/?f=x_A%3D-x_B%5C%5C%5C%5Cy_B%3D6x_A)
Substituting in ![x_{B}^{2}+y_{B}^{2}=30.25](https://tex.z-dn.net/?f=x_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%3D30.25)
![\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904](https://tex.z-dn.net/?f=%5Cleft%20%28-x_%7BA%7D%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%286x_%7BA%7D%20%5Cright%20%29%5E%7B2%7D%3D30.25%5C%5C%5C%5C37x_%7BA%7D%5E2%3D30.25%5C%5C%5C%5Cx_%7BA%7D%3D0.904)
So we have
![\vec{A}=0.904\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3D0.904%5Chat%7Bi%7D)
Magnitude of vector A = 0.904