Question

A circular loop of wire 1.0 cm in radius carries a current of 40 A. The magnetic field at the center of the loop is

Answer 1

The magnetic field strength at the center of a circular loop of wire with a radius of 1.0 cm carrying a 40 A current is calculated using the magnetic field equation for a current loop, which results in a value of 0.002 T or 2 mT.

Step-by-step explanation:

The question deals with the magnetic field strength at the center of a circular loop of wire carrying a current. The standard equation to calculate this magnetic field (B) is given as B = (µ0 * I) / (2 * R), where µ0 is the magnetic constant (also known as the permeability of free space), I is the current through the loop, and R is the radius of the loop. For a loop with a radius of 1.0 cm (0.01 m) and carrying a current of 40 A, the magnetic field at the center can be calculated by inserting the values into the equation: B = (4π * 10-7 T*m/A * 40 A) / (2 * 0.01 m), which simplifies to B = 0.002 T or 2 mT (milliteslas).

This equation derives from the Biot-Savart law and Ampère's law and is specifically for the case of a circular current loop. In general, this formula applies to ideal loops; effects of the wire's thickness and material, as well as the loop's environment, can also affect the actual magnetic field produced.

Answer 2

The magnetic field at the center of the loop is 2.51 × 10⁻³ T

Step-by-step explanation:

The magnetic field at the center of a circular loop is given by

B = μ₀I/2r

Where B is the magnetic field strength in Teslas (T)

μ₀ is the permeability of free space

μ₀ = 4π ×10⁻⁷ N/A²

I is the current in Amperes (A)

and r is the radius of the loop in meters (m)

From the question,

r = 1.0 cm

Convert this to meter (m)

1.0 cm = 1.0 × 10⁻² m = 0.01 m

∴ r = 0.01 m

I = 40 A

Hence, the magnetic field at the center of the loop is

B = μ₀I/2r

B = (4π ×10⁻⁷ × 40) / (2 × 0.01)

B = 5.0265 × 10⁻⁵ / 0.02

B = 2.51 × 10⁻³ T

Hence, the magnetic field at the center of the loop is 2.51 × 10⁻³ T