Question

Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add are called Helmholtz coils. A feature of Helmholtz coils is that the resultant magnetic field between the coils is very uniform. Let R = 11.0 cm, I = 17.0 A, and N = 300 turns for each coil. Place one coil in the y-z plane with its center at the origin and the other in a parallel plane at R = 11.0 cm. Calculate the resultant field Bx at x1 = 2.8 cm, x2 = 5.5 cm, x3 = 7.3 cm, and x4= 11.0 cm.

Answer 1

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$

When x = 5.5 cm, $B_{x2} = 0.0209 T$

when x = 7.3 cm, $B_{x3} = 0.0169 T$

When x = 11.0 cm, $B_{x4} = 0.0103 T$

Explanation:

According to Biot-Savart law,

$B_{x} = \frac{N \mu_{o}IR^{2} }{2(x^{2} +R^{2} )^{3/2} }$.......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

$\mu_{o} = 4\pi * 10^{-7} N/A^{2}$

When x₁ = 2.8 cm = 0.028 m

$B_{x1} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.028^{2} +0.11^{2} )^{3/2} }\\B_{x1} = 0.0265 T$

When x₂ = 5.5cm = 0.055 m

$B_{x2} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.055^{2} +0.11^{2} )^{3/2} }\\B_{x2} = 0.0209 T$

When x₃ = 7.3 cm = 0.073 m

$B_{x3} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.073^{2} +0.11^{2} )^{3/2} }\\B_{x3} = 0.0169 T$

When X₄ = 11.0 cm = 0.11 m

$B_{x4} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.11^{2} +0.11^{2} )^{3/2} }\\B_{x4} = 0.0103 T$