212k views
5 votes
Two long straight wires enter a room through a door. One carries a current of 3.0A into the room while the other carries a current of 5.0A out. The magnitude of the path integral of the magnetic field B around the door frame is:

a.2.5ᴇ−6 T · m
b.3.8ᴇ−6 T · m
c.6.3ᴇ−6 T · m
d.1.0ᴇ−5 T · m
e.none of these

User Eeyore
by
8.6k points

1 Answer

3 votes

Final answer:

The magnitude of the path integral of the magnetic field B around the door frame, given two wires with opposite currents, can be calculated using Ampère's law. The correct answer is 'a. 2.5ᴇ−6 T ⋅ m', after accounting for the net current of 2.0A exiting the room and applying the permeability of free space.

Step-by-step explanation:

The question is asking to find the magnitude of the path integral of the magnetic field B around the door frame, given two wires entering a room with opposite currents. According to Ampère's law, the net magnetic field around a closed loop is proportional to the net current enclosed by that loop. The key here is that one wire is carrying a current of 3.0A into the room, and another wire is carrying 5.0A out of the room. The net current is therefore 5.0A - 3.0A = 2.0A coming out of the room.

Using Ampère's law – ∫ B ⋅ dl = μ0 I where μ0 is the permeability of free space (μ0 = 4π x 10-7 T⋅m/A) and I is the net current (I = 2.0A) – we find the path integral of B over the path around the door frame:

∫ B ⋅ dl = (4π x 10-7 T⋅m/A) ⋅ (2.0A)

This equals 8π x 10-7 T⋅m, which is 2.5 x 10-6 T⋅m after converting π to approximately 3.14. Therefore, the correct answer is 'a. 2.5ᴇ−6 T ⋅ m'.

User Antekone
by
7.8k points