Question
Use the worked example above to help you solve this problem. A certain solenoid consists of 132 turns of wire and has a length of 11.5 cm.
(a) Find the magnitude of the magnetic field inside the solenoid when it carries a current of 0.580A.
T
(b) What is the momentum of a proton orbiting inside the solenoid in a circle with a radius of 0.025 m? The axis of the solenoid is perpendicular to the plane of the orbit.
kg · m/s
(c) Approximately how much wire would be needed to build this solenoid? Assume the solenoid's radius is 6.85 cm.
m
(D)
Suppose you have a 34.5 m length of copper wire. If the wire is wrapped into a solenoid 0.240 m long and having a radius of 0.0571 m, how strong is the resulting magnetic field in its center when the current is 12.4 A?
B = T
5. 1.05/5 points | Previous Answers SerCP11 19.AE.009 EXAMPLE 19.9 The Magnetic Field Inside a Solenoid GOAL Calculate the magnetic field of a solenoid from given data and the momentum of a charged particle in this field PROBLEM A certain solenoid consists of 100 turns of wire and has a length of 10.0 cm . (a) Find the magnitude of the magnetic field inside the solenoid when it carries a current of 0.500 A. (b) What is the momentum of a proton orbiting inside the solenoid in a circle with a radius of 0.020 m? The axis of the solenoid is perpendicular to the plane of the orbit. (c) Approximately how much wire would be needed to build this solenoid? Assume the solenoid's radius is 5.00 cm STRATEGY In part (a) calculate the number of turns per meter and substitute that and given information to get the magnitude of the magnetic field. Part (b) is an application of Newton's second law SOLUTION (A) Find the magnitude of the magnetic field inside the solenoid when it carries a current of 0.500 A Calculate the number of turns per unit N 100 turns 0.100 m1.00 x 103 turns/ Substitute n and I to find the magnitude of the magnetic field: B = 0n1 = (4x 10-7 T·m/A)(1.00 × 103 turns/m)(0.500 A) = 6.28 x 10 -4 (B) Find the momentum of a proton orbiting in a circle of radius 0.020 m near the center of the solenoid Write Newton's second law for the proton Substitute the centripetal acceleration 2 2 a=v.lr: -19 4 Cancel one factor of v on both sides and multiply by r, getting the momentum mv mv-roB-(0.020 m)( 1.60 x 10- C)(6.28 x 10- T) p = mv = 2.01 x 10-24 kg . m/s 10-24 (C) Approximately how much wire would be needed to build this solenoid? Length of wire (number of turns)(2Tr) Multiply the number of turns by the circumference of one loop 2 (1.00% 10 turns)(2n. 0.0500 m)-31.4 m LEARN MORE REMARKS An electron in part (b) would have the same momentum as the proton, but a much higher speed. It would also orbit in the opposite direction. The length of wire in part (c) is only an estimate because the wire has a certain thickness, slightly increasing the size of each loop. In addition, the wire loops aren't perfect circles because they wind slowly up along the solenoid QUESTION What would happen to the instantaneous trajectory of the orbiting proton if the solenoid were suddenly re-oriented so its axis becomes vertical? (Select all that apply aThe proton, as one possibility, could move upward out of the solenoid It depends on the proton's instantaneous position and direction of motion in its orbit when the change occurs The proton would move in a plane oriented at about 45° away from horizontal a The proton, as one possibility, could keep moving in a circle in the horizontal plane The proton would stop moving
Explanation / Answer
given
N = 132 turns
L = 11.5 cm = 0.115 m
a) I = 0.58 A
we know, B = mue*N*I/L
= 4*pi*10^-7*132*0.58/0.115
= 8.37*10^-4 T
b) we know, r = m*v/(B*q)
r = P/(B*q)
==> P = B*q*r
= 8.37*10^-4*1.6*10^-19*0.025
= 3.35*10^-24 kg.m/s
c) length of the wire needed = N*2*pi*r
= 132*2*pi*0.0685
= 56.8 m
D) L = 0.24 m
no of turns on the solenoid, N = 34.5/(2*pi*0.0571)
= 96 turns
A = 12.4 A
so, B = mue*N*I/L
= 4*pi*10^-7*96*12.4/0.24
= 6.23*10^-3 T
