Question

Consider two parallel coaxial circular coils of equal radius R,and the number of turns N carrying equal currents in the same direction , and separated by a distance R. Show that the field on the axis around the midpoint between the coil is uniform over a distance that is small compared to R ,amd it is given by B=0.72 uoNI,/R

Answer 1

Answer:

Explanation:

Explanation:On the axis of the circular coil of radius R and N turns carrying a current I at a distance x from center the magnetic field is given by

formula given in the image attachment.  

 At a distance x = R/2

B =  0.36 uoNI/R

Both are having same current,

Hence, NET MAGNETIC FIELD IS 2B

So, 2B = 0.72uoNI/R

See attachment

Answer 2

Given that,

Radius = R

Number of turns = N

Distance = 2R

Current = I

We need to calculate the magnetic field at midpoint due to loop 1

Using formula magnetic field

$B'=\dfrac{\mu_{0}NIR^2}{2(R^2+x^2)^{\frac{3}{2}}}$

Where, R = radius of circular coil

x = distance

Put the value into the formula

$B'=\dfrac{\mu_{0}NIR^2}{2(R^2+(\dfrca{R}{2})^2)^{\frac{3}{2}}}$

$B'=\dfrac{\mu_{0}NIR^2}{2.7R^3)}$

$B'=\dfrac{\mu_{0}NI}{2.7R}$

We need to calculate the magnetic field at midpoint due to loop 2

Using formula magnetic field

$B''=\dfrac{\mu_{0}NIR^2}{2(R^2+x^2)^{\frac{3}{2}}}$

Where, R = radius of circular coil

x = distance $2(1+\dfrac{1}{4})^{\dfrac{3}{2}}$

Put the value into the formula

$B'=\dfrac{\mu_{0}NIR^2}{2(R^2+(\dfrca{R}{2})^2)^{\frac{3}{2}}}$

$B'=\dfrac{\mu_{0}NIR^2}{2.7R^3)}$

$B'=\dfrac{\mu_{0}NI}{2.7R}$

We need to calculate the net magnetic field

Using formula for net magnetic field

$B=B'+B''$

Put the value into the formula

$B=\dfrac{\mu_{0}NI}{2.7R}+\dfrac{\mu_{0}NI}{2.7R}$

$B=\dfrac{2\mu_{0}NI}{2.7R}$

$B=0.72\dfrac{\mu_{0}IN}{R}$

Hence, The net magnetic field is $0.72\dfrac{\mu_{0}NI}{R}$

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Topic : magnetic field

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