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2 years ago

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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. mastering physics

1 answer:

tensa zangetsu [6.8K]2 years ago

5 0

Answer:

The number of turns is $N = 1750 \ turns$

Explanation:

From the question we are told that

The inner radius is $r_i = 12.0 \ cm = 0.12 \ m$

The outer radius is $r_o = 15.0 \ cm = 0.15 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic field is $B = 3.75 mT = 3.75 *10^{-3} \ T$

The distance from the center is $d = 14.0 \ cm = 0.14 \ m$

Generally the number of turns is mathematically represented as

$N = \frac{2 * \pi * d * B}{ \mu_o * r_o }$

Generally $\mu_o$ is the permeability of free space with value

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

So

$N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }$

$N = 1750 \ turns$

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