Question

Example # 24.1: Find the magnetic field 5 cm from a long wire carrying 12 A. Example # 24.2: Find the magnetic field inside a solenoid with 15,000 turns that is 20 cm long and carries 6.8 A of current. Example # 24.3: Find the force on a proton moving to the right (horizontally) at 330,000 m/s where there is an upward magnetic field of 0.0058 T. Example # 24.4: Find the speed of electrons moving horizontally in the same field as problem 24.3 if they move in a circle of radius 12.0 cm.

Answer 1

24.1) The magnetic field due to a straight current carrying wire at a distance of d form the wire is, В 1,1 2nd (41x10-Tm/A)(12 A) 27 (5.0 x 10-2 m) = 4.8x10-T

24.2) The expression for the magnetic field inside the solenoid is, B=u.nl Here, I is the current in the solenoid, n is the number of turns per unit length of the solenoid, and He is the permeability of free space. The expression for number of turns per unit length is, N n=- Here, N is the number of loops, and I is the length of the solenoid. Combine n= , and B = wonl. B=4(^) = (47x10Im a) (10)(682) =0.6048 T