A solenoid is 3.0cm long and has a radius of 0.50cm. It is wrapped with 500turns of wire carrying a current of 2.0A. The magnetic field at the center of the solenoid is

Explanation

The formula for the magnetic field (\(B\)) at the center of a solenoid is given by:\[ B = \mu_0 \cdot n \cdot I \]Where:- \(\mu_0\) = permeability of free space (\(4\pi \times 10^{-7} \, \text{Tm/A}\))- \(n\) = number of turns per unit length- \(I\) = current passing through the solenoid### Given Data:- Length of the solenoid (\(l\)) = 3.0 cm = \(0.03 \, \text{m}\)- Radius of the solenoid (\(r\)) = 0.50 cm = \(0.005 \, \text{m}\)- Number of turns (\(N\)) = 500- Current (\(I\)) = 2.0 AFirst, let's calculate the number of turns per unit length (\(n\)) of the solenoid:\[ n = \frac{N}{l} \]\[ n = \frac{500}{0.03 \, \text{m}} \]Now, using the formula for the magnetic field at the center of the solenoid:\[ B = \mu_0 \cdot n \cdot I \]\[ B = (4\pi \times 10^{-7} \, \text{Tm/A}) \times \left(\frac{500}{0.03 \, \text{m}}\right) \times 2.0 \, \text{A} \]Let's compute the result:\[ B = 4\pi \times 10^{-7} \times \frac{500 \times 2.0}{0.03} \, \text{T} \]\[ B = 4\pi \times 10^{-7} \times 33333.33 \, \text{T} \]\[ B = 4.19 \times 10^{-2} \, \text{T} \]### Answer:(d) \(4.2 \times 10^{-2} \, \text{T}\)### Conclusion:The accurate magnetic field at the center of the solenoid is \(4.2 \times 10^{-2} \, \text{T}\) (option d), derived from the given data and the solenoid magnetic field formula. Options (a), (b), and (c) do not align with the calculated value.