Answer:
a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H
Explanation:
a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.
Also, Li = Ф where L = inductance of solenoid.
So, Li = NAB
= NA(μ₀ni)
= NA(μ₀Ni/l)
Li = μ₀AN²i/l
dividing both sides by i, we have
So, L = μ₀AN²/l
b. The self- inductance, L = μ₀AN²/l where
A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m
So, L = μ₀AN²/l
L = μ₀πd²N²/4l
L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)
L = 11,103.3 × 10⁻¹¹ H
L = 1.11033 × 10⁻⁷ H
L ≅ 1.11 × 10⁻⁷ H
Answer:
Incomplete question
This is the complete question
For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter
Explanation:
Given the magnetic field
B=2T
Lenght of rod is 1mm
L=1/1000=0.001m
Diameter of rod=1.5mm
d=1.5/1000=0.0015m
Radius is given as
r=d/2=0.0015/2
r=0.00075m
Area of the circle is πr²
A=π×0.00075²
A=1.77×10^-6m²
Given that the voltage applied is 100mV
V=0.1V
Given that resistive is 0.6 Ωm
We can calculate the resistance of the cylinder by using
R= ρl/A
R=0.6×0.001/1.77×10^-6
R=339.4Ω
Then the current can be calculated, using ohms law
V=iR
i=V/R
i=0.1/339.4
i=2.95×10^-4 A
i=29.5 mA
The force in a magnetic field of a wire is given as
B=μoI/2πR
Where
μo is a constant and its value is
μo=4π×10^-7 Tm/A
Then,
B=4π×10^-7×2.95×10^-4/(2π×0.00075)
B=8.43×10^-8 T
Then, the force is given as
F=iLB
Since B=2T
F=iL(2B)
F=2.95×10^-4×2×8.34×10^-8
F=4.97×10^-11N
N2 = 3*n1
T2 = 2*T1
V1 = V2
(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1
Since P2 is 6P1, it is 6 times greater than original pressure
