A solenoid magnet designed for a fusion reactor is wound with superconducting ni

Question

A solenoid magnet designed for a fusion reactor is wound with superconducting niobium-tin (Nb3Sn) wire at a density of 180 turns per meter. The wire carries a current of 30 kA. The magnetic field B inside the solenoid is uniform, whereas outside the solenoid it drops abruptly to zero. The magnetic field at a point half-way through the surface, that is, the field experienced by very current that is creating it, is B/2. What is the direction and magnitude of the magnetic pressure exerted on the coils by the field that they themselves are producing?

Explanation / Answer

n= no of turns per unit length = 180

I = current = 30 kA

B= solenoid magnetic field= u0*n*I

u0= permeability of free space = 4*pi*10-7

B= 4*pi*10-7*180*30*103 = 6.782 T

PB = magnetic pressure = B2 / 2*u0

PB = (6.782)2 / 2*4*pi*10-7

= 18.31*106 Pa

It will be perpendicular to the surface.

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