Answer:
The correct answer is option c. r = R / √2.
Step-by-step explanation:
To find the radius at which the magnitude of the magnetic field inside the wire is half of its maximum value, we can use Ampere's law for a long cylindrical wire. Ampere's law states that the line integral of the magnetic field around a closed path is equal to the product of the enclosed current and the permeability of free space (μ0).
The formula for the magnetic field inside a long cylindrical wire carrying a current density J(r) is given by:
B(r) = (μ0 * J0 * r^2) / (2 * R)
Where:
B(r) = Magnetic field at radial distance r from the wire axis.
μ0 = Permeability of free space (constant ≈ 4π × 10^(-7) Tm/A).
J0 = Constant current density.
r = Radial distance from the wire axis.
R = Radius of the wire.
To find the radius (r) at which the magnetic field is half of its maximum value, we need to set B(r) equal to half of the maximum value of the magnetic field (B_max) and solve for r.
B_max = B(R) = (μ0 * J0 * R^2) / (2 * R) = (μ0 * J0 * R) / 2
Half of the maximum value (B_max/2) is:
B_max/2 = (μ0 * J0 * R) / 4
Now, we set B(r) equal to B_max/2 and solve for r:
(μ0 * J0 * r^2) / (2 * R) = (μ0 * J0 * R) / 4
Simplifying and solving for r:
r^2 = R^2 / 2
r = R / √2
Therefore, the correct answer is option c. r = R / √2.