Question

Ratio of magnetic fields at 10 cm and 20 cm from a infinitely long current carrying wire is​

Answer 1

Answer:

1/2

Explanation:

Answer 2

Answer:

the ratio of magnetic fields at 10 cm and 20 cm from an infinitely long current-carrying wire is​ 2: 1.

Explanation:

We know that the magnetic field of an infinitely long current-carrying wire(B) is​

  $B = \frac{\mu_oI}{2\pi r}$   Where B is the magnitude of the magnetic field, r is the distance from the wire where the magnetic field is calculated, and I is the applied current.

The magnetic field at 10 cm is :

$B_{10} = \frac{\mu_o I}{2\pi (10)}$...(1)

The magnetic field at 20 cm is :

$B_{20} = \frac{\mu_o I}{2\pi (20)}$.....(2)

Now dividing equation (1) with (2)

$\frac{B_{10}}{B_{20}} = \frac{\frac{\mu_o I}{2\pi \times10} }{\frac{\mu_o I}{2\pi \times 20} }$

$\implies \frac{B_{10}}{B_{20}} = \frac{20}{10} = \frac{2}{1}$

Therefore, the ratio of magnetic fields at 10 cm and 20 cm from an infinitely long current-carrying wire is​ 2: 1.

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