- C For Example 22 7 Calculating Field Strength Inside A Solenoid In Section 22 9 Ampere S Law Of The Textbook Repl 1 (12.59 KiB) Viewed 23 times
- C For Example 22 7 Calculating Field Strength Inside A Solenoid In Section 22 9 Ampere S Law Of The Textbook Repl 2 (12.59 KiB) Viewed 23 times
- C For Example 22 7 Calculating Field Strength Inside A Solenoid In Section 22 9 Ampere S Law Of The Textbook Repl 3 (38.15 KiB) Viewed 23 times
EXAMPLE 22.7 n 22.28 Calculating Field Strength inside a Solenoid What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current? Strategy To find the field strength inside a solenoid, we use B = uont. First, we note the number of loops per unit length is N 2000 1000 m1 = 10 cm! 2.00 m Solution Substituting known values gives B = wont = (41 x 10-7Tm/A) (1000 m-') (1600 A) = 2.01 T. Discussion This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000 loops squeezed into a meter's length would produce significant heating. Higher currents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields. 22.29