Answer:
0.047 cm
Explanation:
Parameters given:
Current density, J = 390 A/cm²
Current, I = 0.67A
Current Density, J, is given as:
J = I / A
Area, A, is given as:
A = pi * r² = pi * d²/4
Where d = diameter of loop
J = I / (pi * d²/4)
J = (4 * I) / (pi * d²)
Making d² subject of formula:
d² = (4 * I) / (pi * J)
d² = (4 * 0.67) / (3.142 * 390)
d² = 0.0021871
=> d = 0.047 cm
Answer:
0.0468 cm or 4.68×10⁻⁴ m
Explanation:
Current Density = Current/Cross sectional area.
J = I/A.................... Equation 1
Where J = current density, I = current, A = Cross sectional area of the wire.
Since the wire is cylindrical in shape,
A = πd²/4................ Equation 2
Where d = diameter of the cylindrical wire, π = pie
Substitute equation 2 into equation 1
J = 4I/πd²
Make d the subject of the equation
d = √(4I/Jπ)................ Equation 3
Given: I = 0.67 A, J = 390 A/cm², π = 3.14
Substitute into equation 3
d = √[(4×0.67)/(390×3.14)]
d = √(0.002188)
d = 0.0468 cm.
Hence the diameter of the wire = 0.0468 cm or 4.68×10⁻⁴ m
