Consider a cross-section of an electromagnet having an air-gap of 5 cm as shown in the figure. It consists of a magnetic material \( \mu = 20000 \mu_0 \) and is driven by a coil having \( NI = 10^4 \), where \( N \) is the number of turns and \( I \) is the current in Ampere.
Ignoring the fringe fields, the magnitude of the magnetic field \( \vec{B} \) (in Tesla, rounded off to two decimal places) in the air-gap between the magnetic poles is \(\underline{\hspace{2cm}}\).
Show Hint
To calculate the magnetic field in a material with an air-gap, use Ampere's law and account for the length of the air-gap.
The magnetic field in the air-gap can be calculated using Ampere's law:
\[
B = \frac{\mu N I}{l},
\]
where:
- \( \mu = 20000 \mu_0 \),
- \( N I = 10^4 \),
- \( l = 5 \, \text{cm} = 0.05 \, \text{m} \) is the length of the air-gap.
First, calculate \( \mu_0 \), the permeability of free space, which is \( 4\pi \times 10^{-7} \, \text{H/m} \). Now, substituting the values:
\[
B = \frac{20000 \times 4\pi \times 10^{-7} \times 10^4}{0.05} \approx 0.24 \, \text{Tesla}.
\]
Thus, the magnitude of the magnetic field is approximately \( 0.24 \, \text{Tesla} \).
