Question

At a distance of 2.5 cm from a long straight current-carrying wire, the magnetic field has a value of 1.3 T. What would be the value of the magnetic field at a point that is twice as far from the wire, when the wire carries twice as much current

Answer 1

The magnetic field is
`B_2 = 1.3 \ T`

Step-by-step explanation:

From the question we are told that

The distance from the wire is d = 2.5 cm = 0.25 m

The magnetic field is B = 1.3 T

Generally the magnetic field that is as a result of as current carrying conductor is mathematically represented as

`B = (\mu_o * I)/(2 \pi d)`

so

`1.3 = (\mu_o * I)/(2 \pi d)`

given that the distance and the current is doubled

`B_2 = (\mu_o * 2I)/(2 \pi (2d) )`

=>
`B_2 = (\mu_o * I)/(2 \pi d)`

Hence

`B_2 = 1.3 = (\mu_o * I)/(2 \pi d)`

=>
`B_2 = 1.3 \ T`