Solution :
a). B at the center :
[tex]$=\frac{u\times I}{2R}$[/tex]
Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.
Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE
b). Also, the sum of the fields must be zero.
Therefore,
[tex]$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$[/tex]
So,
[tex]$\frac{I_1}{d_1}= \frac{I_2}{d_2}$[/tex]
[tex]$=\frac{16}{21}=\frac{I_2}{32}$[/tex]
[tex]$I_2=24.38 $[/tex] A
Therefore, the current in the outer wire is 24.38 ampere.
Answer:
(a) counter clockwise
(b) 24.38 A
Explanation:
inner diameter, d = 21 cm
inner radius, r = 10.5 cm
Current in inner loop, I = 16 A clock wise
Outer diameter, D = 32 cm
Outer radius, R = 16 cm
(a) The magnetic filed due to the inner wire is inwards to the plane of paper. According to the Maxwell's right hand thumb rule, the direction of magnetic field in outer wire should be outwards so that the net magnetic field is zero at the center.
So, the direction of current in outer wire is counter clock wise in direction.
(b) Let the current in outer wire is I'.
The magnetic field due to the inner wire is balanced by the magnetic field due to the outer wire.
[tex]\frac{ \mu 0}{4\pi}\times \frac{2 I}{r}=\frac{\mu 0}{4\pi}\times \frac{2 I'}{R}\\\frac{16}{10.5}=\frac{I'}{16}\\\\I' = 24.38 A[/tex]
