Magnetic force on a wire inside magnetic field is given by formula
![F = i (L X B)here we will haveℓ = (1.382 i − 2.095 j) mB = (−0.350 j + 0.550 k) t.i = 5.52 Anow as per above formula we have[tex]F = 5.52 * (1.382 i − 2.095 j) X (−0.350 j + 0.550 k)](https://tex.z-dn.net/?f=F%20%3D%20i%20%28L%20X%20B%29%3C%2Fp%3E%3Cp%3Ehere%20we%20will%20have%3C%2Fp%3E%3Cp%3E%E2%84%93%20%3D%20%281.382%20i%20%E2%88%92%202.095%20j%29%20m%3C%2Fp%3E%3Cp%3EB%20%3D%20%28%E2%88%920.350%20j%20%2B%200.550%20k%29%20t.%3C%2Fp%3E%3Cp%3Ei%20%3D%205.52%20A%3C%2Fp%3E%3Cp%3Enow%20as%20per%20above%20formula%20we%20have%3C%2Fp%3E%3Cp%3E%5Btex%5DF%20%3D%205.52%20%2A%20%281.382%20i%20%E2%88%92%202.095%20j%29%20X%20%28%E2%88%920.350%20j%20%2B%200.550%20k%29)


So magnetic force on the wire is given by above expression with all three components shown with i, j, k.
x component of force is 6.36 N
y component of force is 4.196 N
z component of force is 2.67 N
Answer:
9.7 x 10¹¹ .
Explanation:
2.5 % of 110 W = 2.75 J/s
energy of one photon
= hc / λ
=
= .036 x 10⁻¹⁷ J
No of photons emitted
= 2.75 / .036 x 10⁻¹⁷
= 76.38 x 10¹⁷
Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance
photons passing per unit area of this sphere
= 76.38 x 10¹⁷ / 4π ( 2.8)²
Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing
= 
= 9.7 x 10¹¹ .
Answer:
a. Zin = 41.25 - j 16.35 Ω
b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c. Pin = 216 w
d. PL = Pin = 216 w
e. Pg = 478.4 w , Pzg = 262.4 w
Explanation:
a.
Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]
βl = 2π / λ * 0.15 λ = 54 °
Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]
Zin = 41.25 - j 16.35 Ω
b.
I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶
V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)
V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c.
Pin = ¹/₂ * Re * [V₁ * I₁]
Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)
Pin = 216 w
d.
The power PL and Pin are the same as the line is lossless input to the line ends up in the load so
PL = Pin
PL = 216 w
e.
Pg Generator
Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)
Pg = 478.4 w
Pzg dissipated
Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50
Pzg = 262.4 w
To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,


Here,
m = Mass
V = Velocity
Replacing,


Therefore the final kinetic energy of the two car system is 72.6kJ
sorry Idk the answer ..???