Question

If the identical coils have radii of 1.17 cm and are 2.31 cm apart, with 38 turns of wire apiece, what current should they both carry to produce a magnetic field of 4.40 10-5 T halfway between them

Answer 1

0.0133 A

Step-by-step explanation:

We are given that

Radius of coil=r=1.17 cm=
`(1.17)/(100)=0.0117 m`

1 m=100 cm

Distance between the coil=d=2.13 cm=0.0213 m

`x=(d)/(2)=(0.0213)/(2)=0.01065 m`

Number of turns=N=38

2 N=2(38)=76

Magnetic field produced in coils=B=
`4.4* 10^(-5) T`

Magnetic field along the axis of symmetry of loop wire

`B=(2N)*\frac{\mu_0 Ir^2}{(x^2+r^2)^{(3)/(2)}}`

Substitute the values

`4.4* 10^(-5)=76\frac{(4\pi* 10^(-7)* I* (0.117)^2}{(0.01065)^2+(0.0117)^2)^{(3)/(2)}}`

`I=4.4* 10^(-5)*\frac{((0.01065)^2+(0.0117)^2)^{(3)/(2)}}{76* 4\pi* 10^(-7)* (0.117)^2}=0.0133 A`

I=0.0133 A