Question

A long solenoid with 11.2 turns/cm and a radius of 9.74 cm carries a current of 24.2 mA. A current of 17.1 A exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis in centimeters will the direction of the resulting magnetic field be at 62.3° to the axial direction? (b) What is the magnitude of the magnetic field there?

Answer 1

Answer with Explanation:

We are given that

Number of turns =n=11.2 /cm=$11.2\times 100=1120/m$

1 m=100 cm

I=24.2 mA=$24.2\times 10^{-3} A$

$1m A=10^{-3} A$

Radius of solenoid,r=9.74 cm=$9.74\times 10^{-2} m$

Current in straight conductor,I'=17.1 A

a.Magnetic field in solenoid=$B_s=\,u_0 nI=4\pi \times 10^{-7}\times 1120\times 24.2\times 10^{-3}=3.4\times 10^{-5} T$

Magnetic field in straight wire=$B'=\frac{\mu_0I'}{2\pi r'}=\frac{2\times 10^{-7}\times 17.1}{r'}=\frac{34.2\times 10^{-7}}{r'}$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

$\theta=62.3^{\circ}$

$\frac{B'}{B}=tan 62.3^{\circ}$

$\frac{34.2\times 10^{-7}}{r'\times 3.4\times 10^{-5}}=1.9047$

$r'=\frac{34.2\times 10^{-7}}{3.4\times 10^{-5}\times 1.9047}=0.053m$

b.Magnitude of magnetic field=$\sqrt{B^2+B'^2}$

$\sqrt{(3.4\times 10^{-5})^2+(\frac{2\times 10^{-7}\times 17.1}{0.053})^2}= 7.29\times 10^{-5} T$

Hence,the magnitude of magnetic field =$7.29\times 10^{-5} T$