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

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Euphemia 1 year 2021-09-04T06:03:46+00:00 2 Answers 331 views 0

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    0
    2021-09-04T06:04:53+00:00

    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:

    • v is the velocity (2.0i+4.0j+6.0k) m/s
    • F is the magnetic force (0.4i-20j+12k) N
    • B is the magnetic force (ai+bj+ck) T
    • q is the charge 2 C

    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!

    0
    2021-09-04T06:05:12+00:00

    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

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