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

Answer:

The magnetic field is  $B_2 = 1.3 \ T$

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

         $B = \frac{\mu_o * I}{2 \pi d}$

so  

      $1.3 = \frac{\mu_o * I}{2 \pi d}$

given that the distance and the current is doubled

         $B_2 = \frac{\mu_o * 2I}{2 \pi (2d) }$

=>      $B_2 = \frac{\mu_o * I}{2 \pi d}$

Hence  

        $B_2 = 1.3 = \frac{\mu_o * I}{2 \pi d}$

=>      $B_2 = 1.3 \ T$