Respuesta :
Answer:
Induced emf in first coil is 0.986 [tex]\mu T[/tex] and in second case 0.396 [tex]\mu T[/tex]
Explanation:
Given:
Number of turns per centimeter [tex]n = 40[/tex]
Current [tex]I = 0.240[/tex] A
Current rate [tex]\frac{dI}{dt} = \frac{0.240}{0.70} = 0.343[/tex] [tex]\frac{A}{s}[/tex]
The magnetic field in solenoid is given by,
[tex]B = \mu _{o} nI[/tex]
Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex]
We write,
[tex]\frac{dB}{dt} = \mu_{o} n \frac{dI}{dt}[/tex]
[tex]\frac{dB}{dt} = 4\pi \times 10^{-7} \times 40 \times 0.343[/tex]
[tex]\frac{dB}{dt} = 172.3 \times 10^{-7}[/tex]
(A)
Number of turns [tex]N = 48[/tex]
Radius of coil [tex]r = \frac{d}{2} = 1.95 \times 10^{-2}[/tex] m
From faraday's law
[tex]\epsilon = NA \frac{dB}{dt}[/tex]
Where [tex]A = \pi r^{2} = 3.14 (1.95 \times 10^{-2} ) ^{2} = 11.93 \times 10^{-4}[/tex] [tex]m^{2}[/tex]
[tex]\epsilon = 48 \times 11.93 \times 10^{-4} \times 172.3 \times 10^{-7}[/tex]
[tex]\epsilon = 98665.87 \times 10^{-11}[/tex]
[tex]\epsilon = 0.986 \mu T[/tex]
(B)
Number of turns [tex]N = 96[/tex]
Radius of coil [tex]r = \frac{d}{2} = 3.9 \times 10^{-2}[/tex] m
From faraday's law
[tex]\epsilon = NA \frac{dB}{dt}[/tex]
Where [tex]A = \pi r^{2} = 3.14 (3.9 \times 10^{-2} ) ^{2} = 47.76 \times 10^{-4}[/tex] [tex]m^{2}[/tex]
[tex]\epsilon = 48 \times 47.96 \times 10^{-4} \times 172.3 \times 10^{-7}[/tex]
[tex]\epsilon = 396648.38 \times 10^{-11}[/tex]
[tex]\epsilon = 0.396 \mu T[/tex]
Therefore, induced emf in first coil is 0.986 [tex]\mu T[/tex] and in second case 0.396 [tex]\mu T[/tex]
