A Loop wire has a diameter of 1.00 cm. It carries a current which generates a magnetic field of 3.00 MT at the center of the loop.
What is the current in the loop?
To find the current in the loop, we can use the formula for the magnetic field produced by a current-carrying loop at its center:
B = (μ₀ * I * N) / (2 * R)
Where:
- B is the magnetic field magnitude
- μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A)
- I is the current flowing through the wire
- N is the number of turns in the loop
- R is the radius of the loop
In this case, we are given the magnetic field magnitude (B) as 3.00 MT (T is tesla, and M stands for mega = 10⁶), and the diameter of the loop wire, which implies the radius (R) is half of that diameter.
First, let's convert the diameter of the loop wire from centimeters to meters:
1.00 cm * (1 m / 100 cm) = 0.01 m
Since the radius (R) is half of the diameter, we can use the converted value for the radius.
Using the formula above, we can rearrange it to solve for current (I):
I = (2 * R * B) / (μ₀ * N)
Plugging in the values we know:
R = 0.01 m
B = 3.00 MT = 3.00 × 10⁶ T (since Mega = 10⁶)
μ₀ = 4π × 10⁻⁷ T·m/A
Now let's solve for the current (I):
I = (2 * 0.01 m * 3.00 × 10⁶ T) / (4π × 10⁻⁷ T·m/A)
I = (0.02 * 3.00 × 10⁶) / (4π × 10⁻⁷) A
I = 6.00 × 10⁸ / 1.26 × 10⁻⁶ A
I ≈ 4.76 × 10¹⁴ A
Therefore, the current in the loop is approximately 4.76 × 10¹⁴ Amperes.