Question

A coil of 100 turns is wound uniformly over a
iron ring of 0.5 meter length and carry a
current of 1 Amp, what will be the magnetic
field strength of the magnet
Options
O 20 AT/m
50 AT/m
O 100 AT/m
O
200 AT/m
Clear Response​

Answer 1

Answer:

According to Bio-sawart Law magnetic flux density is given by B

$b = \frac{uni}{2r} = \frac{4\pi \times {10}^{ - 7} \times 100 \times 1}{2 \times 0.5} \\ = 1.257 \times \times {10}^{ - 4}$

Answer 2

Answer:

The magnetic field strength, H, calculated is $200A/m$.

Therefore. option d) $200A/m$ is correct.

Step-by-step explanation:

Given,

The number of turns of the coil, N = $100turns$

The length of the iron ring, L = $0.5m$

The current carried by the iron ring, I = $1A$

The strength of the magnetic field, H =?

Now, as we know,

• Magnetic field strength, H = $\frac{N\times I}{L}$

After putting the values of turns, length, and current in the equation, we get:

• H = $\frac{100\times 1}{0.5}$
• H = $200A/m$

Hence, the magnetic field strength, H, calculated is $200A/m$.