A solenoid is formed by winding 11.0 m of insulated silver(resistivity 1.59 x 10

Question

A solenoid is formed by winding 11.0 m of insulated silver(resistivity 1.59 x 10-8 ·m) wire around ahollow cylinder. The turns are wound as closely as possible withoutoverlapping, and the insulating coat on the wire is negligiblythin. When the solenoid is connected to an ideal (no internalresistance) 3.00-V battery, the magnitude of the magnetic fieldinside the solenoid is found to be 4.69 x 10-3 T.Determine the radius of the wire. Hint: Because the solenoid isclosely coiled, the number of turns per unit length depends uponthe radius of the wire.

Explanation / Answer

The resistance of the wire = L/A
                                      = *L/( r2 )
So the current I = V/R                          = V* r2 / *L

To determine the number of turns per unit length first calculatethe length of wire per turn N = 2R                 where R is the radius of the solenoid  
So the number of turns N = L/(2R* 2r) the number of turns per length n = 1/(2R*2r)

Now the Magnetic field B = µo*n*I                          B = 4.69*10-3 T
                     B = µo/(4R*r)*V**r^2/*L

simplifying we have B = µo*V*r^2/(4*R*r**L)

                          r = 4**R*L*B/µo*V)                                 = 4 * 1.59*10-8 m * 11.0 m *   4.69*10-3 T / 4 *10-7 * 3.00 V                              = 8.70*10-4 m

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