Explanation:
It is given that,
Length of wire, l = 0.53 m
Current, I = 0.2 A
(1.) Approximate formula:
We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m
The formula for magnetic field at some distance from the wire is given by :
![B=\dfrac{\mu_oI}{2\pi r}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20r%7D)
![B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%200.2%5C%20A%7D%7B2%5Cpi%20%5Ctimes%200.02%5C%20m%7D)
B = 0.000002 T
![B=10^{-5}\ T](https://tex.z-dn.net/?f=B%3D10%5E%7B-5%7D%5C%20T)
(2) Exact formula:
![B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20r%7D%5Cdfrac%7Bl%7D%7B%5Csqrt%7Bl%5E2%2B4r%5E2%7D%20%7D)
![B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_o%5Ctimes%200.2%5C%20A%7D%7B2%5Cpi%20%5Ctimes%200.02%5C%20m%7D%5Ctimes%20%5Cdfrac%7B0.53%5C%20m%7D%7B%5Csqrt%7B%280.53%5C%20m%29%5E2%2B4%280.02%5C%20m%29%5E2%7D%20%7D)
B = 0.00000199 T
or
B = 0.000002 T
Hence, this is the required solution.
Answer:
![q_2 = - 1.66\times 10^{-13} c](https://tex.z-dn.net/?f=q_2%20%3D%20-%201.66%5Ctimes%2010%5E%7B-13%7D%20c)
Explanation:
Given data:
if two charge are opposite in charge then force will be attractive between then or vice versa
if one charge is positive then other charge will be negative
from coulomb's law
![f = \frac{1\times q_1\times q_2}{4\pi \epsilon_o \times r^2}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%5Ctimes%20q_1%5Ctimes%20q_2%7D%7B4%5Cpi%20%5Cepsilon_o%20%5Ctimes%20r%5E2%7D)
![\frac{1}{4\pi \epsilon_o} = 9\times 10^{9}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%20%3D%209%5Ctimes%2010%5E%7B9%7D)
![3.1 N = 9\times 10^{9} \frac{91.4\times q_2}{0.23^2}](https://tex.z-dn.net/?f=3.1%20N%20%3D%209%5Ctimes%2010%5E%7B9%7D%20%5Cfrac%7B91.4%5Ctimes%20q_2%7D%7B0.23%5E2%7D)
![q_2 = \frac{3.1\times 0.21^2}{9\times 10^{9}\times 91.4}](https://tex.z-dn.net/?f=q_2%20%3D%20%5Cfrac%7B3.1%5Ctimes%200.21%5E2%7D%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%2091.4%7D)
![q_2 = - 1.66\times 10^{-13} c](https://tex.z-dn.net/?f=q_2%20%3D%20-%201.66%5Ctimes%2010%5E%7B-13%7D%20c)
Answer:
Percentage change in tension is 3.8%
Explanation:
We have given initially frequency
= 440 Hz
Let tension in the string at this frequency is ![T_1](https://tex.z-dn.net/?f=T_1)
Now second frequency is ![f_2=448.4Hz](https://tex.z-dn.net/?f=f_2%3D448.4Hz)
Frequency in string is given by
![f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2L%7D%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%20%7D%7D)
From the relation we can say that
![\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bf_1%7D%7Bf_2%7D%3D%5Csqrt%7B%5Cfrac%7BT_1%7D%7BT_2%7D%7D)
![{\frac{T_2}{T_1}}=1.038](https://tex.z-dn.net/?f=%7B%5Cfrac%7BT_2%7D%7BT_1%7D%7D%3D1.038)
Percentage change in tension is equal to
%
So percentage change in tension is 3.8%
Answer:
<h2>2 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
![a = \frac{f}{m} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bf%7D%7Bm%7D%20%20%5C%5C%20)
f is the force
m is the mass
From the question we have
![a = \frac{2000}{1000} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B2000%7D%7B1000%7D%20%20%5C%5C%20)
We have the final answer as
<h3>2 m/s²</h3>
Hope this helps you
Answer:
it ends when clouds above start to break apart. Some tornadoes only last seconds. Others can last much longer. They come in many shapes and sizes.