Answer:
a) [tex]\vec F_{B} = 8.766\times 10^{-14}\,T\,k[/tex], b) [tex]\vec F_{B} = -8.766\times 10^{-14}\,T\,k[/tex]
Explanation:
a) The magnetic force experimented by a particle has the following vectorial form:
[tex]\vec F_{B} = q\cdot \vec v \times \vec B[/tex]
The charge of the electron is equal to [tex]-1.602\times 10^{-19}\,C[/tex]. Then, cross product can be solved by using determinants:
[tex]\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}[/tex]
The magnetic force is:
[tex]\vec F_{B} = 8.766\times 10^{-14}\,T\,k[/tex]
b) The charge of the proton is equal to [tex]1.602\times 10^{-19}\,C[/tex]. Then, cross product has the following determinant:
[tex]\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}[/tex]
The magnetic force is:
[tex]\vec F_{B} = -8.766\times 10^{-14}\,T\,k[/tex]
This question involves the concepts of the magnetic field, triple-scalar product, and magnetic force.
(a) The magnitude of the magnetic force on the electron is " 0.876 x 10⁻¹³ N".
(b) The magnitude of the magnetic force on the proton is "-0.876 x 10⁻¹³ N".
(a)
The magnitude of a magnetic force applied on a charged particle due to the magnetic field is given by the following formula:
[tex]F=q.v\ x\ B[/tex]
where,
F = magnetic force = ?
q = charge on electron = -1.6 x 10⁻¹⁹ (i + j + k) C
v = velocity = (2.6 i + 3.2 j +0 k) x 10⁶ m/s
B = magnetic field = (0.041 i - 0.16 j + 0 k) T
Therefore,
[tex]F = (-1.6\ x\ 10^{-19}(i+j+k)\ C)).(2.6\ x\ 10^6i+3.2\ x\ 10^6j+0k)\ x\ (0.041i-0.16j+0k)[/tex]
This can be solved using the scalar-triple product:
[tex]F = \left|\begin{array}{ccc}-1.6\ x\ 10^{-19}&-1.6\ x\ 10^{-19}&-1.6\ x\ 10^{-19}\\2.6\ x\ 10^6&3.2\ x\ 10^6&0\\0.041&-0.16&0\end{array}\right|[/tex]
Expanding by column 3, we get:
[tex]F = (-1.6\ x\ 10^{-19})[(2.6\ x\ 10^6)(-0.16)-(3.2\ x\ 10^6)(0.041)]\\F = (-1.6\ x\ 10^{-19})(-0.416\ x\ 10^{6}-0.131\ x\ 10^6)\\F = (-1.6\ x\ 10^{-19})(-0.547\ x\ 10^{6})\\[/tex]
F = 0.876 x 10⁻¹³ N
(b)
For the proton the charge changes to:
q = 1.6 x 10⁻¹⁹ (i + j + k) C
Therefore,
[tex]F = (1.6\ x\ 10^{-19}(i+j+k)\ C)).(2.6\ x\ 10^6i+3.2\ x\ 10^6j+0k)\ x\ (0.041i-0.16j+0k)[/tex]
This can be solved using the scalar-triple product:
[tex]F = \left|\begin{array}{ccc}1.6\ x\ 10^{-19}&1.6\ x\ 10^{-19}&1.6\ x\ 10^{-19}\\2.6\ x\ 10^6&3.2\ x\ 10^6&0\\0.041&-0.16&0\end{array}\right|[/tex]
Expanding by column 3, we get:
[tex]F = (1.6\ x\ 10^{-19})[(2.6\ x\ 10^6)(-0.16)-(3.2\ x\ 10^6)(0.041)]\\F = (1.6\ x\ 10^{-19})(-0.416\ x\ 10^{6}-0.131\ x\ 10^6)\\F = (1.6\ x\ 10^{-19})(-0.547\ x\ 10^{6})\\[/tex]
F = - 0.876 x 10⁻¹³ N
Learn more about magnetic force here:
https://brainly.com/question/10353944?referrer=searchResults
Attached picture shows the formula of magnetic force.
