Answer :
Answer:
a) B = 0.0151 T
b) [tex] \theta = 4.737 * 10^-^6 Tm^2 [/tex]
c) [tex] E = 7.896*10^-^4 [/tex]
d) [tex] = 60.4 T [/tex]
Explanation:
Given:
Diameter, d = 2.5 cm convert to meters = 0.025 m
Length, L = 30.0 cm convert to meters = 0.3 m
Number of turns, N = 300
Current, I = 12 A
a) Let's use the formula below to find the magnetic field.
[tex] B = \frac{u_0 N I}{L} [/tex]
[tex] B = \frac{4\pi * 10^-^7 * 300 * 12}{0.3} [/tex]
[tex] B = \frac{0.004524}{0.3} [/tex]
B = 0.0151 T
b) Given a radius, r = 1.00 cm, let's convert it to meters = 0.01 m
Let's use the formula below to find magnetic flux.
[tex] \theta = BA [/tex]
where [tex] A = \pi r^2 [/tex]
Therefore,
[tex] \theta = B \pi r^2 [/tex]
[tex] = 0.0151 \pi * (0.01)^2 [/tex]
[tex] \theta = 4.737 * 10^-^6 Tm^2 [/tex]
c) Given that the current in the solenoid uniformly goes from 12 A to 10 A in 0.001 seconds, the EMF generated will be:
[tex]E = \frac{- change in \theta}{change in t} = \frac{-u_0 N}{L} * \pi r^2 * \frac{change in I}{change in t}[/tex]
[tex] E = \frac{-4\pi * 10^-^7 * 300}{0.3} * \pi *(0.01)^2 * \frac{10 - 12}{0.001} [/tex]
[tex] -0.00125663 * \pi *(0.01)^2 * -2000[/tex]
[tex] E = 7.896*10^-^4 [/tex]
d) Let's use the formula:
[tex] \frac{u_m N I}{L} = 4000 \frac{u_0 NI}{L} [/tex]
[tex] 4000 * 0.0151 [/tex]
[tex] = 60.4 T [/tex]