Answer:
168.57 mV
Explanation:
Initial magnetic flux = BA , B magnetic field and A is area of loop
= .35 x 3.14 x .37²
= .15 Weber
Final magnetic flux
= - .2 x 3.14 x .37²
= - .086 Weber
change in flux
.15 + .086
= .236 Weber
rate of change of flux
= .236 / 1.4
= .16857 V
= 168.57 mV
<h2>
Answer:</h2>
143μH
<h2>
Explanation:</h2>
The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
<em>From the question;</em>
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
<em>But;</em>
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
<em>Substitute these values into equation (i) as follows;</em>
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
Answer:
I think the answer 1
Explanation:
im probably wrong too i dont know
