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3 years ago

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A certain superconducting magnet in the form of a solenoid of length 0.24 m can generate a magnetic field of 7.0 T in its core w

hen its coils carry a current of 90 A. The windings, made of a niobium-titanium alloy, must be cooled to 4.2 K. Find the number of turns in the solenoid.

1 answer:

tatiyna3 years ago

5 0

Answer:

$N=14854.5turns$

Explanation:

Given data

Length L=0.24 m

Magnetic field B=7.0 T

Current I=90 A

To find

Number of turns N

Solution

As we know that magnetic field is given as:

$B=u_{o}I\frac{N}{L}$

Rearrange the equation and solve for N

So

$B=u_{o}I\frac{N}{L}\\\frac{B}{u_{o}I} =\frac{N}{L} \\N=\frac{LB}{u_{o}I}\\N=\frac{(0.24m)(7.0T)}{4\pi *10^{-7}*(90A)} \\N=14854.5turns$

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The equation for energy dissipated in the resistor is,

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Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

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Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

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Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

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$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

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