Question

A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform magnetic field it creates inside the coil in μT? Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil (meaning the center of the circular cross section of the coil). What is the magnitude of the total magnetic field created half way between the straight wire and the inner coil walls in μT? Hint: think about each magnetic field's direction.

Answer 1

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

$B=\mu_{0} ni$

Put the value into the formula

$B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}$

$B_{c}=0.0000100\ T$

$B_{c}=10.02\times10^{-6}\ T$

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance $d=\dfrac{5}{2}$

$d=2.5\ cm$

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

$B_{w}=\dfrac{\mu_{0}I}{2\pi d}$

Put the value into the formula

$B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}$

$B_{w}=4.0\times10^{-6}\ T$

This field is perpendicular to the wire.

The magnitude of magnetic field is

$B=\sqrt{B_{c}^2+B_{w}^2}$

$B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}$

$B=0.00001078\ T$

$B=10.78\times10^{-6}\ T$

$B=10.78\ \mu\ T$

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.