Answer:
The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C
Explanation:
Please see the attachments below
Please see below solution:
The electric field just outside its surface is
=
2
,
where is a radius of sphere, is a charge of sphere, =
1
4ε0
.
The electric field is
=
1
4ε0
2 =
1
ε0
∙
=
σ
ε0
,
where = 4
2
is an area of sphere, σ =
Q
S
is a surface charge density.
Answer:
.
Answer:
0.00221 N
Explanation:
Given that,
The charge on the particle,
The speed of the particle, v = 65.8 m/s (+x direction)
Magnetic field, B = 0.545 T (in +y direction) and 0.828 T in the positive z direction.
The magnetic force is given by the formula as follows :
![F=q(v\times B)](https://tex.z-dn.net/?f=F%3Dq%28v%5Ctimes%20B%29)
Substitute all the values,
![F=34\times 10^{-6}\times (65.8i\times (0.545j+0.828 k))\\\\=34\times 10^{-6}\times (65.8i\times 0.545j +65.8i\times 0.828 k)\\\\=34\times 10^{-6}\times(35.86k +(-54.48j))\\\\=34\times 10^{-6}\times \sqrt{35.86^2+54.48^2} \\\\=0.00221\ N](https://tex.z-dn.net/?f=F%3D34%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20%20%2865.8i%5Ctimes%20%280.545j%2B0.828%20k%29%29%5C%5C%5C%5C%3D34%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20%2865.8i%5Ctimes%200.545j%20%2B65.8i%5Ctimes%200.828%20k%29%5C%5C%5C%5C%3D34%5Ctimes%2010%5E%7B-6%7D%5Ctimes%2835.86k%20%2B%28-54.48j%29%29%5C%5C%5C%5C%3D34%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20%5Csqrt%7B35.86%5E2%2B54.48%5E2%7D%20%5C%5C%5C%5C%3D0.00221%5C%20N)
So, the magnitude of the magnetic force on the particle is equal to 0.00221 N.
I believe the answer is <span>216396.148556 meters per second</span>