Answer:
The maximum emf induced in the ring
= (2.882 × 10⁻⁷) V
Explanation:
According to the law of electromagnetic induction, the emf induced in the ring is given by
E = N BA w sin wt
The maximum emf induced is
E = N BA w
B = 30.5 μT = (30.5 × 10⁻⁶) T
A = (πD²/4)
D = 1.75 cm = 0.0175 m
A = (π×0.0175²/4) = 0.000240625 m²
Nw = 2π × 6.25 = 39.29 rad/s
E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29
E = (2.882 × 10⁻⁷) V
Hope this Helps!!!
Answer:
Explanation:
Electric field due to charge at origin
= k Q / r²
k is a constant , Q is charge and r is distance
= 9 x 10⁹ x 5 x 10⁻⁶ / .5²
= 180 x 10³ N /C
In vector form
E₁ = 180 x 10³ j
Electric field due to q₂ charge
= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²
= 30.33 x 10³ N / C
It will have negative slope θ with x axis
Tan θ = .5 / √.5² + .8²
= .5 / .94
θ = 28°
E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j
= 26.78 x 10³ i - 14.24 x 10³ j
Total electric field
E = E₁ + E₂
= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j
= 26.78 x 10³ i + 165.76 X 10³ j
magnitude
= √(26.78² + 165.76² ) x 10³ N /C
= 167.8 x 10³ N / C .
Answer:
(a) the force is 8.876 N
(b) the magnitude of each charge is 4.085 μC
Explanation:
Part (a)
Given;
coulomb's constant, K = 8.99 x 10⁹ N.m²/C²
distance between two charges, r = 10 cm = 0.1 m
force between the two charges, F = 15 N
when the distance between the charges changes to 13 cm (0.13 m)
force between the two charges, F = ?
Apply Coulomb's law;

Part (b)
the magnitude of each charge, if they have equal magnitude

where;
F is the force between the charges
K is Coulomb's constant
Q is the charge
r is the distance between the charges

Hello,
Your answer to this problem is 400/3
Hope this helps!
Answer:
-414.96 N
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration


The force the ground exerts on the parachutist is -414.96 N
If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase