Answer:
The number of turns is ![N = 1750 \ turns](https://tex.z-dn.net/?f=N%20%20%3D%201750%20%5C%20turns%20)
Explanation:
From the question we are told that
The inner radius is ![r_i = 12.0 \ cm = 0.12 \ m](https://tex.z-dn.net/?f=r_i%20%3D%20%2012.0%20%5C%20%20cm%20%20%3D%20%200.12%20%5C%20%20m)
The outer radius is ![r_o = 15.0 \ cm = 0.15 \ m](https://tex.z-dn.net/?f=r_o%20%3D%20%2015.0%20%5C%20%20cm%20%20%3D%20%200.15%20%5C%20%20m)
The current it carries is ![I = 1.50 \ A](https://tex.z-dn.net/?f=I%20%3D%20%201.50%20%5C%20%20A)
The magnetic field is ![B = 3.75 mT = 3.75 *10^{-3} \ T](https://tex.z-dn.net/?f=B%20%20%3D%20%20%203.75%20mT%20%3D%203.75%20%2A10%5E%7B-3%7D%20%5C%20%20T)
The distance from the center is ![d = 14.0 \ cm = 0.14 \ m](https://tex.z-dn.net/?f=d%20%3D%20%2014.0%20%5C%20cm%20%20%3D%20%200.14%20%5C%20%20m)
Generally the number of turns is mathematically represented as
![N = \frac{2 * \pi * d * B}{ \mu_o * r_o }](https://tex.z-dn.net/?f=N%20%20%3D%20%20%5Cfrac%7B2%20%2A%20%20%5Cpi%20%20%2A%20d%20%20%2A%20%20B%7D%7B%20%5Cmu_o%20%2A%20%20r_o%20%7D)
Generally
is the permeability of free space with value
![\mu_o = 4\pi * 10^{-7} \ N/A^2](https://tex.z-dn.net/?f=%5Cmu_o%20%20%3D%20%204%5Cpi%20%2A%2010%5E%7B-7%7D%20%5C%20N%2FA%5E2)
So
![N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }](https://tex.z-dn.net/?f=N%20%20%3D%20%20%5Cfrac%7B2%20%2A%20%203.142%20%20%20%2A%200.14%20%2A%20%203.75%20%2A10%5E%7B-3%7D%20%7D%7B%204%5Cpi%20%2A%2010%5E%7B-7%7D%20%20%2A%200.15%20%20%7D)
![N = 1750 \ turns](https://tex.z-dn.net/?f=N%20%20%3D%201750%20%5C%20turns%20)
Answer:
The value is ![KE_b =0.710 \ J](https://tex.z-dn.net/?f=KE_b%20%3D0.710%20%5C%20J)
Explanation:
From the question we are told that
The mass of the bullet is ![m_b = 5.00 \ g = 0.005 \ kg](https://tex.z-dn.net/?f=m_b%20%20%3D%205.00%20%5C%20g%20%20%3D%200.005%20%5C%20%20kg)
The mass of the wood is ![m_w = 900 \ g = 0.90\ kg](https://tex.z-dn.net/?f=m_w%20%3D%20%20900%20%5C%20%20g%20%20%3D%20%200.90%5C%20%20kg)
The height attained by the combined mass is ![h = 8.0 \ cm = 0.08 \ m](https://tex.z-dn.net/?f=h%20%3D%20%208.0%20%5C%20cm%20%20%3D%20%200.08%20%5C%20m)
Generally according to the law of energy conservation
![KE _b = PE_c](https://tex.z-dn.net/?f=KE%20_b%20%20%3D%20%20PE_c)
Here
is the kinetic energy of the bullet before collision.
and
is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as
![PE_m = [m_b + m_w] * g * h](https://tex.z-dn.net/?f=PE_m%20%20%3D%20%20%5Bm_b%20%20%2B%20m_w%5D%20%2A%20%20g%20%2A%20%20h)
So
![KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08](https://tex.z-dn.net/?f=KE_b%20%3DPE_c%20%20%20%3D%20%5B0.005%20%20%2B%200.90%5D%20%2A%209.8%20%2A0.08)
=> ![KE_b =0.710 \ J](https://tex.z-dn.net/?f=KE_b%20%3D0.710%20%5C%20J)
The sun <u><em>appears</em></u> brighter than any other star.
(It isn't really, but it looks that way because it's much much much much much much closer to us than any other star.)
Answer:
The current through the inductor at the end of 2.60s is 9.7 mA.
Explanation:
Given;
emf of the inductor, V = 41.0 mV
inductance of the inductor, L = 13 H
initial current in the inductor, I₀ = 1.5 mA
change in time, Δt = 2.6 s
The emf of the inductor is given by;
![V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA](https://tex.z-dn.net/?f=V%20%3D%20L%5Cfrac%7Bdi%7D%7Bdt%7D%20%5C%5C%5C%5CV%20%3D%20%5Cfrac%7BL%28I_1-I_o%29%7D%7Bdt%7D%20%5C%5C%5C%5CL%28I_1-I_o%29%20%3D%20V%2Adt%5C%5C%5C%5CI_1-I_o%20%3D%20%5Cfrac%7BV%2Adt%7D%7BL%7D%5C%5C%5C%5CI_1%20%3D%20%20%5Cfrac%7BV%2Adt%7D%7BL%7D%20%2B%20I_o%5C%5C%5C%5CI_1%20%3D%20%5Cfrac%7B41%2A10%5E%7B-3%7D%2A2.6%7D%7B13%7D%20%2B1.5%2A10%5E%7B-3%7D%5C%5C%5C%5CI_1%20%3D%208.2%2A10%5E%7B-3%7D%20%2B%201.5%2A10%5E%7B-3%7D%5C%5C%5C%5CI_1%20%3D%209.7%20%2A10%5E%7B-3%7D%20%5C%20A%5C%5C%5C%5C%20I_1%20%3D%209.7%20%5C%20mA)
Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.
Answer:
the branch of mechanics concerned with the interaction of electric currents with magnetic fields or with other electric currents.
Explanation: