Question

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How many equally spaced turns must it have so that it will produce a magnetic field of 3.75 mT at points within the coils 14.0 cm from its center? Enter your answer numerically.

Answer 1

Answer:

The number of turns, N = 1750

Explanation:

It is given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

The magnetic field at points within the coils 14 cm from its center is, $B=3.75\ mT=3.75\times 10^{-3}\ T$

R = 14 cm = 0.14 m

Current, I = 1.5 A

The formula for the magnetic field at some distance from its center is given by :

$B=\dfrac{\mu_o NI}{2\pi R}$

$N=\dfrac{2\pi R B}{\mu_o I}$

$N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}$

N = 1750

So, the number of turns must have in a toroidal solenoid is 1750. Hence, this is the required solution.

Answer 2

Answer:

The number of turns are 1750.

Explanation:

Given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

Magnetic field, $B=3.75\ mT=3.75\times 10^{-3}\ T$

To find,

It is required to find the number of turns that will produce the magnetic field of 3.75 mT at points within the coils 14.0 cm from its center.

The formula for the magnetic field at some distance from its center is given by :

$B=\dfrac{\mu_o NI}{2\pi R}\\\\N=\dfrac{2\pi R B}{\mu_o I}\\\\N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}$

$N=1750$

So, the number of turns are 1750.