Answer:
I = 4.642 Ampere.
x = 2.145 cm
Explanation:
a) As we know, the magnetic field on the axis of the loop is given as
[tex]B = \frac{ \mu NIa^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }[/tex]
where a = radius of loop
x = point on the axis of loop
N = No of turns of coil
Current in the loop for which the magnetic field at the center is 0.0750 Tesla is given as x = 0
Therefore, the above equation can be rewritten as
[tex]B_{x} = \frac{ \mu NI}{2a}[/tex]
I = [tex]\frac{2aB}{μN}[/tex]
by putting values
I =[tex]\frac{2 X 0.0750 T X 0.028 m}{4\pi X 10^{-7} T.\frac{m}{A} X 720}[/tex]
Current in the loop for which the magnetic field at the center is 0.0750 Tesla = I = 4.642 Ampere
b)
Now for part b the magnetic field at a distance x from the center is given as
[tex]B = \frac{ \mu N I a^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }[/tex]
multiply and divide by a on both sides we get
[tex]B = \frac{ \mu NI}{2a} X \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }[/tex]
As we know, according to Biot sivorts law,the Magnetic Field at the Center of a circular loop is given as
[tex]B = \frac{ \mu NI}{2a}[/tex] ( Magnetic field at center) = [tex]B_{c}[/tex]
So we got magnetic field at any point x as
[tex]B_{x}[/tex] = [tex]B_{c}[/tex] X [tex]\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }[/tex]
For magnetic field at x is half of the B at center
[tex]B_{x}[/tex] = [tex]\frac{1}{2}[/tex] [tex]B_{c}[/tex]
from the above two equations
[tex]\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }[/tex] = [tex]\frac{1}{2}[/tex]
[tex](x^{2} + a^{2})^{3} = 4a^{6}[/tex]
[tex]x = \sqrt{4^{\frac{1}{3}} -1 } X a^{}[/tex]
Putting a = 2.80 cm
We have x = 2.145 cm Ans
μ
If the magnetic field at the center of the coil is 0.0750 Tesla, the current in the coil must be equal to 4.64 Amperes.
Given the following data:
Radius = 2.80 cm to m = 0.028 m.
Number of turns = 720 turns.
Magnetic field = 0.0750 T.
Scientific data:
Permeability of free space = [tex]4\pi \times 10^{-7}[/tex]
In order to determine the current, we would apply Ampere's law of magnetic field.
Mathematically, Ampere's law of magnetic field is given by this formula:
[tex]I=\frac{2B r}{\mu_o N}[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]I=\frac{2 \times 0.0750 \times 0.028 }{4\pi \times 10^{-7} \times 720}\\\\I=\frac{0.0042}{9.05 \times 10^{-4}}[/tex]
Current, I = 4.64 Amperes.
Since the magnetic field is half its value at the center, we have:
[tex]B_c = \frac{1}{2} (\frac{\mu_o NI}{2r} )\\\\[/tex]
Also, the magnetic field at the center of the coil is given by:
[tex]B_C =\frac{\mu_o NIr^2}{2(x^2+r^2)^\frac{3}{2} }[/tex]
Equating the two equations and solving for x, we have:
[tex]\frac{1}{2} (\frac{\mu_o NI}{2r} ) =\frac{\mu_o NIr^2}{2(x^2+r^2)^\frac{3}{2} }\\\\2r^3 = (x^2+r^2)^\frac{3}{2}\\\\(2r^3)^\frac{2}{3} = x^2+r^2\\\\x^2=(2r^3)^\frac{2}{3}-r^2\\\\x^2 = (2\times (0.028)^3)^\frac{2}{3}-0.028^2\\\\x^2 = 0.001245-0.000784\\\\x=\sqrt{0.000461}[/tex]
x = 0.02147 meter or 2.147 centimeter.
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