Question

An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 4.95 cm^2. When the current is 12.5 A, the energy stored is 0.395 J. How many turns does the winding have?

Answer 1

13

hope this helps.........

Answer 2

Answer:

281 turns

Explanation:

Energy stored (U) = 0.395 J

$u \: = \frac{1}{2} l {i}^{2} = 0.395$

After Rearranging, we get

$l(self \: inductance) = \frac{2u}{ {i}^{2} } = 5.05 \times {10}^{ - 3}$

Therefore,

$l = \frac{ \beta {n}^{2}a }{2\pi r} = 5.05 \times {10}^{ - 3}\ H$

Where,

$\beta (permeability \: of \: free \: space) = 4\pi \times {10}^{ - 7}$

$n \: = number \: of \: turns$

Now

$n \: = \sqrt{ \frac{2\pi \: r \: \times l}{ \beta \times a} }$

$n = \sqrt{ \frac{2 \times 3.14 \times15.5 \times 5.05 \times {10 }^{ - 3} }{4\pi \times {10}^{ - 7} \times4.95 } }$

$n = 281.1 (approx.) = 281 turns$