Question

An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.

Answer 1

Answer:

Fₓ = 0,  $F_{y}$ = 0  and  $F_{z}$ = - 3.115 10⁻¹⁵   N

Explanation:

The magnetic force given by the expression

       F = q v xB

the bold are vectors,  the easiest analytical way to determine this force in solving the determinant

   $F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}$

   F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

   F =i^0 + j^0   - k^  3.115 10⁻¹⁵   N

   

Fₓ = 0

$F_{y}$ = 0

$F_{z}$ = - 3.115 10⁻¹⁵   N