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ValentinkaMS [17]
3 years ago
5

Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of

4 A and the other is carrying a current of 9 A. The magnitude of the magnetic field exactly in between the two wires is ____ T.
Physics
1 answer:
mart [117]3 years ago
5 0

Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

    d_1 =  0.3 \  m

B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

    d_2 =  0.3 \  m  

This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

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