Consider a tightly wound 100 turn coil of radius 10 cm carrying a current of 1A. What is the magnitude of magnetic field at the centre of the coil?

Question

Answer

You can solve this by using B=[(mue0)×IN]/2a where •mue0 is the permeability of vacuum(4л×10^-7) •I=1A •N=100 •a=10cm and I think the ans: 6.28×10^-6

Answer

We can solve this by using the derived formula                               B= NUi/2r     ( where U =4pi*10^-7 )by this we get           B = 6.28*10^-4

Answer

6.28×10^4 T

Answer

Magnetic field at the center of a coil carrying current

The magnetic field at the center of a coil carrying current can be calculated using the formula:

B = μ0 * I * N / (2 * R)

Where B is the magnetic field, μ0 is the permeability of free space, I is the current, N is the number of turns in the coil, and R is the radius of the coil.

Calculating the magnetic field

Using the given values, we can calculate the magnetic field at the center of the coil as follows:

B = (4π × 10^-7 T·m/A) * 1A * 100 / (2 * 0.1m) = 1.2566 × 10^-3 T

Therefore, the magnitude of the magnetic field at the center of the coil is 1.2566 × 10^-3 T.

Explanation

The magnetic field at the center of a coil carrying current is a result of the individual magnetic fields produced by each turn of the coil. When current flows through a wire, it creates a magnetic field around the wire. In a coil, the magnetic fields produced by each turn of the coil add up at the center, resulting in a stronger magnetic field.

The strength of the magnetic field depends on the current, the number of turns in the coil, and the radius of the coil. Increasing any of these factors will increase the strength of the magnetic field.

Conclusion

In conclusion, the magnetic field at the center of a tightly wound 100 turn coil of radius 10 cm carrying a current of 1A is 1.2566 × 10^-3 T. This can be calculated using the formula B = μ0 * I * N / (2 * R). The magnetic field is a result of the individual magnetic fields produced by each turn of the coil.

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