Share
A 2.0 c charge moves with a velocity of (2.0i+4.0j+6.0k)m/s and experiences a magnatic force of (0.4i-20j+12k)N. The x component of the magn
Question
A 2.0 c charge moves with a velocity of (2.0i+4.0j+6.0k)m/s and experiences a magnatic force of (0.4i-20j+12k)N. The x component of the magnatic field is equal to zero. Determine the y component of the magnatic field
in progress
0
Physics
1 year
2021-09-04T06:03:46+00:00
2021-09-04T06:03:46+00:00 2 Answers
331 views
0
Answers ( )
Answer:
[tex]\vec{B}_{y}=6T[/tex]
Explanation:
Here we can use the Lorentz force equation.
[tex]\vec{F}_{B}=q(\vec{v}\times \vec{B})[/tex]
We know:
So we will have:
[tex] (0.4i-20j+12k)=2((2.0i+4.0j+6.0k) \times (ai+bj+ck))[/tex]
Let’s solve the cross product, knowing that x component of B is 0, it means a=0.
[tex] (0.4i-20j+12k)=2((2.0i+4.0j+6.0k) \times (0i+bj+ck))[/tex]
[tex] (0.4i-20j+12k)=2((4c-6b)i-2cj+2bk)[/tex]
Comparing the k component we have:
[tex]12=2b [/tex]
[tex]b=6 [/tex]
If we see b is the y-component of the magnetic field, therefore [tex]\vec{B}_{y}=6T[/tex]
I hope it helps you!
Answer:
-0.033 units
Explanation:
According to Lorentz force law, the magnetic force, F, on a moving charge, q, moving with a velocity, v, in a magnetic field, B, is given by;
F = q v x B —————-(i)
Where;
F, v and B are vectors. Therefore, equation (i) represents a vector product of the velocity and magnetic field vectors.
From the question;
v = (2.0i + 4.0j + 6.0k)m/s
F = (0.4i – 20j + 12k)N
q = 2.0C
Let the magnetic field vector be given by;
B = ai + bj + ck ———————(*)
Where;
a, b and c are the magnitudes of the x, y and z components of the magnetic field.
Substitute the values of F, v, B and q into equation (i) as follows;
(0.4i – 20j + 12k) = 2.0(2.0i + 4.0j + 6.0k) x (ai + bj + ck)
Expanding the second bracket gives
(0.4i – 20j + 12k) = (4.0i + 8.0j + 12.0k) x (ai + bj + ck) —————(ii)
Solving the right hand side of equation (ii) which is the vector product gives;
| |
| i j k |
(0.4i – 20j + 12k) = | 4.0 8.0 12.0 |
| a b c |
| |
(0.4i – 20j + 12k) = (8.0c – 12.0b)i – (4.0c -12.0a)j + (4.0b – 8.0a)k —-(iii)
Comparing both sides of equation (iii) gives the following three equations;
0.4 = 8.0a – 12.0b ——————–(iv)
-20 = 4.0c – 12.0a ———————(v)
12 = 4.0b – 8.0a ———————-(vi)
From the question, it is given that the x component of the magnetic field is equal to zero. Now, recall that from equation (*) above, the magnitude of the x-component of the magnetic field is given as a
Therefore;
a = 0
To get the y component, which is b, substitute the value of a = 0 into equation (iv) as follows;
0.4 = 8.0(0) – 12.0b
0.4 = 0 – 12.0b
0.4 = – 12.0b
Solve for b;
b = [tex]\frac{-0.4}{12.0}[/tex] = -0.033
Therefore the y component of the magnetic field is -0.033 units