P31_073.PDF

73. Letting the current in solenoid 1 be i , we calculate the ﬂux linkage in solenoid 2. The mutual inductance,

then, is this ﬂux linkage divided by i . The magnetic ﬁeld inside solenoid 1 is parallel to the axis and has

uniform magnitude B = µ 0 in 1 ,where n 1 is the number of turns per unit length of the solenoid. The

cross-sectional area of the solenoid is πR 1 .Since B is normal to the cross section, the ﬂux here is

Φ= AB = πR 1 µ 0 n 1 i.

Since the magnetic ﬁeld is zero outside the solenoid, this is also the ﬂux through a cross section of

solenoid 2. The number of turns in a length of solenoid 2 is N 2 = n 2 , and the ﬂux linkage is

N 2 Φ= n 2 πR 1 µ 0 n 1 i.

The mutual inductance is

M = N 2 Φ

= πR 1 µ 0 n 1 n 2 .

M does not depend on R 2 because there is no magnetic ﬁeld in the region between the solenoids.

Changing R 2 does not change the ﬂux through solenoid 2, but changing R 1 does.