A segment of wire of length L is along the x axis centered at x=0. Which of the following is a correct integral expression for the magnetic field at point P (centered on the wire segment at y=b) due to the current I flowing left to right in the segment of length L? In all answers below the limits of integration are from -L/2 to L/2. a. μ0I/4π∫ dx b/(b2 + x2)3/2 kb. μ0I/4π∫ dx b/(b2 + x2)3/2 j c. μ0I/4π∫ dx /(b2 + x2) kd. -μ0I/4π∫ dx /(b2 + x2)1/2 ke. none of 1-5

b. μ0I/4π∫ dx b/(b2 + x2)³/² j

Explanation:

 The wire of length L centered at origin ( x =0 and y=0 ) carries current of I . We have to find magnetic field at point ( x = 0 , y = b ) .

First of all we shall consider magnetic field due to current element idx which is at x distance away from origin . magnetic field

dB =

component of magnetic field along y- axis at point P

where θ is angle between y - axis and dE .

component of magnetic field along y- axis at point P

The same magnetic field will also exist due to current element dx at x distance away on negative x - axis

The perpendicular component will cancel out .

This is magnetic field dE due to small current element

Magnetic field due to whole wire

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

the results is a decrease in air temperature with an increase of altitude.

explanation: