At first, we have:
q = 1.24 × 1 0 − 8 C ; v ⃗ = [ 1.49 × 1 0 1 , − 3.85 × 1 0 1 , 0 , ] m s q=1.24\times10^{-8}\,C; \,\vec{v}=\begin{bmatrix}
1.49\times 10^1, & -3.85\times 10^1, & 0,
\end{bmatrix}
\tfrac{m}{s} q = 1.24 × 1 0 − 8 C ; v = [ 1.49 × 1 0 1 , − 3.85 × 1 0 1 , 0 , ] s m
( a ) B ⃗ = [ 1.40 , − 2.30 , 0 , ] T (a)\,\vec{B} = \begin{bmatrix}
1.40, & -2.30, & 0,
\end{bmatrix} T ( a ) B = [ 1.40 , − 2.30 , 0 , ] T
( b ) B ⃗ = [ − 1.25 , 2.00 , − 3.50 , ] T (b)\,\vec{B} = \begin{bmatrix}
-1.25, & 2.00, & -3.50,
\end{bmatrix}T ( b ) B = [ − 1.25 , 2.00 , − 3.50 , ] T
F ⃗ = q ⋅ v ⃗ × B ⃗ \vec{F} = q\cdot \vec{v}\times \vec{B} F = q ⋅ v × B
Using the definition of the Lorentz force will allow us to calculate the force exerted by each magnetic field, thus we find that:
(a) F B ⃗ = ( 1.24 × 1 0 − 8 ) ( i ^ j ^ k ^ 14.9 − 38.5 0 1.40 − 2.30 0 ) N \vec{F_{B}}=(1.24\times10^{-8}) \begin{pmatrix}
\,\widehat{i} & \widehat{j} & \widehat{k}\,\\
14.9 & -38.5 & 0\\
1.40 &-2.30 & 0
\end{pmatrix}\,N F B = ( 1.24 × 1 0 − 8 ) ⎝ ⎛ i 14.9 1.40 j − 38.5 − 2.30 k 0 0 ⎠ ⎞ N
F B ⃗ = ( 1.24 × 1 0 − 8 N ) ( 14.9 − 38.5 1.40 − 2.30 ) k ^ \vec{F_{B}}=(1.24\times10^{-8}\,N)\begin{pmatrix}
14.9 & -38.5\\
1.40 &-2.30
\end{pmatrix}\widehat{k} F B = ( 1.24 × 1 0 − 8 N ) ( 14.9 1.40 − 38.5 − 2.30 ) k
F B ⃗ = ( 1.24 ∗ 1 0 − 8 N ) ( ( − 14.9 ∗ 2.30 ) − ( − 1.40 ∗ 38.5 ) ) k ^ \vec{F_{B}}=(1.24*10^{-8}\,N)((-14.9*2.30)-(-1.40*38.5))
\widehat{k} F B = ( 1.24 ∗ 1 0 − 8 N ) (( − 14.9 ∗ 2.30 ) − ( − 1.40 ∗ 38.5 )) k
( a ) F B ⃗ = ( 2.43 × 1 0 − 7 N ) k ^ (a) \vec{F_{B}}=(2.43\times10^{-7}\,N)\,\widehat{k} ( a ) F B = ( 2.43 × 1 0 − 7 N ) k
Then, for b:
(b) F B ⃗ = ( 1.24 × 1 0 − 8 ) ( i ^ j ^ k ^ 14.9 − 38.5 0 − 1.25 − 2.00 − 3.50 ) N \vec{F_{B}}=(1.24\times10^{-8}) \begin{pmatrix}
\widehat{i} & \widehat{j} & \widehat{k}\\
14.9 & -38.5 & 0\\
-1.25 &-2.00 & -3.50
\end{pmatrix}\,N F B = ( 1.24 × 1 0 − 8 ) ⎝ ⎛ i 14.9 − 1.25 j − 38.5 − 2.00 k 0 − 3.50 ⎠ ⎞ N
( 1.24 ) ( 1 0 − 8 ) N ) [ ( 14.9 − 38.5 − 1.25 − 2.00 ) k ^ + ( i ^ j ^ 14.9 − 38.5 ) ( − 3.50 ) ] (1.24)(10^{-8})\,N)\left [ \begin{pmatrix}
14.9 & -38.5 \\
-1.25 & -2.00
\end{pmatrix}\widehat{k}\,+\begin{pmatrix}
\widehat{i} & \widehat{j} \\
14.9 & -38.5
\end{pmatrix}(-3.50)
\right ] ( 1.24 ) ( 1 0 − 8 ) N ) [ ( 14.9 − 1.25 − 38.5 − 2.00 ) k + ( i 14.9 j − 38.5 ) ( − 3.50 ) ]
=( 1.24 ∗ 1 0 − 8 ) [ ( − 29.8 − 48.125 ) k ^ + ( − 38.5 i ^ − 14.9 j ^ ) ( − 3.50 ) ] (1.24*10^{-8})[(-29.8-48.125)\widehat{k}\,+(-38.5\widehat{i}-14.9\widehat{j})(-3.50)] ( 1.24 ∗ 1 0 − 8 ) [( − 29.8 − 48.125 ) k + ( − 38.5 i − 14.9 j ) ( − 3.50 )]
= ( 134.75 i ^ + 52.15 j ^ − 77.925 k ^ ) ( 1.24 ∗ 1 0 − 8 ) N (134.75\widehat{i}+52.15\widehat{j}-77.925\widehat{k})(1.24*10^{-8})\,N ( 134.75 i + 52.15 j − 77.925 k ) ( 1.24 ∗ 1 0 − 8 ) N
(b) F B ⃗ = ( 16.71 i ^ + 6.47 j ^ − 9.66 k ^ ) ∗ 1 0 − 7 N \vec{F_{B}}=(16.71\widehat{i}+6.47\widehat{j}-9.66\widehat{k}\,) *10^{-7}\,N F B = ( 16.71 i + 6.47 j − 9.66 k ) ∗ 1 0 − 7 N
In conclusion, the forces exerted on the charged particle due to the magnetic field are:
( a ) F B ⃗ = ( 2.43 ∗ 1 0 − 7 N ) k ^ (a)\,\vec{F_{B}}=(2.43*10^{-7}\,N)\,\widehat{k} ( a ) F B = ( 2.43 ∗ 1 0 − 7 N ) k ( b ) F B ⃗ = ( 16.71 i ^ + 6.47 j ^ − 9.66 k ^ ) ∗ 1 0 − 7 N (b)\,\vec{F_{B}}=(16.71\widehat{i}+6.47\widehat{j}-9.66\widehat{k}\,) *10^{-7}\,N ( b ) F B = ( 16.71 i + 6.47 j − 9.66 k ) ∗ 1 0 − 7 N
Reference:
Young, H. D., & Freedman, R. A. (2014). Sears and Zemansky's University Physics (with Modern Physics Technology Update) . Edinburgh: Pearson. 
#332078 on Oct 2022
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