\(magnetic field intensity,
H=NI/l\)
A solenoid has 500 turns of wire and a length of 0.4 meters. If a current of 2 A flows through the solenoid, calculate the magnetic field strength inside the solenoid.
Hint: Use the formula for the magnetic field inside a solenoid:
H=NILH = \frac{NI}{L}H=LNI
Where:
- HHH is the magnetic field strength (in A/m),
- NNN is the number of turns,
- III is the current (in amperes),
- LLL is the length of the solenoid (in meters).
