TheThe smallest diameter of wrapper that will fit the candy bar is 5cm.
<h3>Smallest diameter</h3>
Given:
AD = DC
AC = AD + DC
Segment AD = 2.5cm
Hence:
AC = AD + AD (AD = DC)
AC = 2AD
Where:
AD=2.5
So,
AC = 2(2.5cm)
AC = 5cm
Therefore the smallest diameter of wrapper that will fit the candy bar is 5cm.
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2+2-1 that's three quick maths.
Answer: 3.61×10^5 A
Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.
The formulae is given below as
B = u×Ia²/2(x²+a²)^3/2
Where B = strength of magnetic field on the axis of a circular loop = 4.15T
u = permeability of free space = 1.256×10^-6 mkg/s²A²
I = current on loop =?
a = radius of loop.
Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m
x = distance of the sensor away from center of loop = 2.10 cm = 0.021m
By substituting the parameters into the formulae, we have that
4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2
4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2
4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)
4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I
I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9
I = 4.80×10^-4 / 1.32665 ×10^-9
I = 3.61×10^5 A
