Strategic Intervention Material for Ampere's Law

1. STRATEGIC INTERVENTION MATERIAL
AMPERE’S LAW
Prepared by:
Alvin P. Cajiles
Teacher I
Narra National High School

2. OBJECTIVE
Calculate magnetic fields for
highly symmetric current
configurations using
Ampere’s Law
(STEM12_GPEM-IIIi-65).

3. GUIDE CARD
Gauss’s Law allowed us to
find the net electric field
due to any charge
distribution by applying
symmetry.
Carl Friedrich
Gauss
Are we able to find the net
magnetic field due to
current if there is
symmetry? Let’s get help
from Ampere’s Law.
André-Marie
Ampère

4. GUIDE
CARD
Ampere’s Law is a
useful law that relates
the net magnetic field
along a closed loop to
the electric field current
passing through the
loop.
First discovered by
André-Marie Ampère in

5. GUIDE CARD
The integral around a closed path of the
component of the magnetic field tangent to the
direction of the path is equals to µ times the
current I intercepted by the area within the
path.
𝐵. 𝑑𝑠 = 𝜇0 𝐼𝑒𝑛𝑐

6. GUIDE CARD
𝐵. 𝑑𝑠 = 𝜇0 𝐼𝑒𝑛𝑐
line integral
B.ds is integrated around
a closed loop called
Amperian Loop
Permeability
of free space
(constant)
Enclosed
current by the
curve
AMPERE’S LAW

7. GUIDE CARD
IMPORTANT NOTES
- All currents have to be steady and do not
change with time.
-Only currents crossing the area inside the path
are taken into account.
-Current have to be taken with their algebraic
sign by using the Right Hand Rule

8. GUIDE CARD
THE RIGHT HAND RULE for DETERMINING
THE DIRECTION OF MAGNETIC FIELD

9. ACTIVITY CARD No.1
Find the magnetic field outside a long straight
wire with current I and radius r.
I
r
Amperian
Loop
Wire surface
Direction of
Integration
Magnetic field
B
ds
r
cross-section of
the wire

10. ACTIVITY CARD No.2
What is the direction of magnetic field due to a
current passing through a solenoid?
Remember to use the
Right Hand Rule
where the thumb
takes the direction of
the current.

11. ACTIVITY CARD No.3
Use Ampere’s Law to
determine the
magnetic field
strength outside a
solenoid with n turns
(coils) per unit
length.
L

12. ASSESSMENT CARD No.1
Find the magnetic field inside a current-
carrying wire.
ds
r
R
R
I B
Amperian
Loop

13. ASSESSMENT CARD No.2
Calculate the magnetic field B in a
wire with radius 1µm and the
current passing through it is 1 A.
Note that 𝝁 𝟎 = 𝟒𝝅 𝒙 𝟏𝟎−𝟕
𝑻. 𝒎/𝑨.

14. ASSESSMENT CARD No.3
The magnetic field strength of a solenoid
is 0.0270T. Its radius is 0.40 m and length is
0.40 m. How many turns are there in the
solenoid if the steady current passing
through it is 12.0 A?

15. ENRICHMENT CARD
Magnetic Field of a
Toroid
The current enclosed by the dashed
line is just the number of loops times
the current in each loop. Ampere’s
Law then gives the magnetic field by
𝐵 2𝜋𝑟 = 𝜇0 𝑁𝐼
𝐵 =
𝜇0 𝑁𝐼
2𝜋𝑟

16. ENRICHMENT CARDToroid is a useful device used in many applications in telecommunication,
music instruments, medical field, ballasts, EMI filter among others to
direct and restrict magnetic fields.

17. ENRICHMENT CARD
Additional Helpful Reading
www.learnapphysics.com/apphysicsc/magnetism.php
AP Physics C – Ampere’s Law
https://www.youtube.com/watch?v=pLyrVDJ3qas&t=5s

18. REFERENCE CARD
AP Physics C – Amperes Law. (2013). Retrieved from https://www.youtube.com/watch?v=pLyrVDJ3qas&t=5s
Elert, G. (n.d.). Ampere’s Law – Problems – The Physics Hypertextbook. Retrieved from https://physics.info/law-
ampere/problems/shtml.
SS: Magnetic field due to current in a straight wire. (2015). Retrieved from https://www.miniphysics.com/ss-
magnetic-field-due-to-current-in-a-straight-wire.html
Solenoid. (n.d.). Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html
Toroidal Magnetic Field. (n.d.). Retrieved from http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/toroid.html
White, R. (n.d.). Learn AP Physics – Magnetism. Retrieved from
http://www.learnapphysics.com/apphysicsc/magnetism.php

20. ACTIVITY No.1
𝐵. 𝑑𝑠 = 𝜇0 𝐼
𝐵 𝑑𝑠 = 𝜇0 𝐼
𝐵 2𝜋𝑟 = 𝜇0 𝐼
𝑑𝑠 = 2𝜋𝑟 → 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒
𝐵 =
𝜇0 𝐼
2𝜋𝑟

21. ANSWER KEY
Activity No.2: Remember that Right Hand
Rule says that the thumb
takes the direction of the
current and the curl of the
palm is the direction of the
magnetic field. Thus, in this
case, the magnetic field is
like passing through the
solenoid.

22. ANSWER KEY
Activity No.3:
𝑩. 𝒅𝒔 = 𝝁 𝟎 𝑰
𝑩 𝒅𝒔 = 𝝁 𝟎 𝑰
𝒅𝒔 = 𝑳 → 𝒕𝒐𝒕𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒊𝒓𝒆(𝒔𝒐𝒍𝒆𝒏𝒐𝒊𝒅)
𝑰 → 𝒏𝑰 𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒖𝒓𝒏𝒔
𝑩 𝑳 = 𝝁 𝟎 𝒏𝑰
∴ 𝑩 =
𝝁 𝟎 𝒏𝑰
𝑳
→ 𝒎𝒂𝒈𝒏𝒆𝒕𝒊𝒄 𝒇𝒊𝒆𝒍𝒅 𝒊𝒏 𝒂 𝒔𝒐𝒍𝒆𝒏𝒐𝒊𝒅

23. ANSWER KEY
Assessment No.1:
𝐵. 𝑑𝑠 = 𝜇0 𝐼
𝐵 𝑑𝑠 = 𝜇0 𝐼
𝐵 2𝜋𝑟 = 𝜇0 𝐼
𝑑𝑠 = 2𝜋𝑟 → 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒
𝐼𝑒𝑛𝑐 =
𝜋𝑟2 𝑖
𝜋𝑅2 →charge enclosed is proportional to the area encircled by the loop
𝐵 (2𝜋𝑟) = 𝜇0
𝜋𝑟2 𝑖
𝜋𝑅2
∴ 𝐵 =
𝜇0 𝐼𝑟
2𝜋𝑅2 → 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑖𝑛 𝑎 𝑐𝑜𝑎𝑥𝑖𝑎𝑙 𝑐𝑎𝑏𝑙𝑒

24. ANSWER KEY
Assessment No.2:
𝐵 =
(4𝜋 𝑥 10−7
𝑇. 𝑚/𝐴)(1𝐴)
(2𝜋)(1 𝑥 10−3 𝑚)
𝐵 = 200 𝜇𝑇

25. ANSWER KEY
Assessment No.3:
𝐵 =
𝜇0 𝑁𝐼
𝐿
𝑁 =
𝐵𝐿
𝜇0 𝐼
𝑁 =
𝐵𝐿
𝜇0 𝐼
=
(0.0270 𝑇)(0.40 𝑚)
(4𝜋 𝑥 10−7 𝑇.𝑚/𝐴)(12 𝐴)
= 716 𝑡𝑢𝑟𝑛𝑠

26. ONLINE VERSION
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