The light of that star actually gives off
2 answers:
You might be interested in
Answer:
current in loops is 52.73 μA
Explanation:
given data
side of square a = b = 2.40 cm = 0.024 m
resistance R = 1.20×10^−2 Ω
edge of the loop c = 1.20 cm = 0.012 m
rate of current = 120 A/s
to find out
current in the loop
solution
we know current formula that is
current = voltage / resistance .................a
so current = 1/R × d∅/dt
and we know here that
flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b
so
d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c
so from equation a we get here current
current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt
current = ( 4π××0.024 / 2π(1.20×
) × ln (0.024 + 0.012/0.012) × 120
solve it and we get current that is
current = 4 ×× 1.09861 × 120
current = 52.73 × A
so here current in loops is 52.73 μA
<h3>Given, </h3>
Force,F = 4000 N
Area,a = 50 m²
<h3>We know that, </h3>Pressure = Force/Area
★ Putting the values in the above formula,we get: