A fighter plane of length 20 m, wing span (distance from tip of one...

${\text{V}}_{\text{B}} = {\text{B}}_{H} \left(5\right) \left(2 4 0\right)$

${\text{B}}_{\text{H}} = \text{B cos} \theta , ....$

${B}_{v} = Bsin\theta$

${\text{B}}_{\text{H}} = \dfrac{5 \sqrt{5} \times 1 {0}^{- 5}}{3}$

${\text{B}}_{\text{v}} = \dfrac{1 0}{3} \times 1 {0}^{- 5} \text{T}$

${\text{V}}_{\text{B}} = \dfrac{5 \sqrt{5}}{3} \times 1 {0}^{- 5} \times 5 \times 2 4 0$

${\text{V}}_{\text{B}} = \text{44.6} mV = 4 5 \text{mV}$

${\text{V}}_{\text{w}} = {\text{B}}_{\text{v}} ℓ \text{V}$$= \dfrac{10}{3} \times {10}^{- 5} \times 15 \times 240$

$= 1 {0}^{- 4} \times 1 2 0 0$

${\text{V}}_{\text{w}} = 1 2 0 mV$

(left side at fighter voltage)