Magnetic Effects of Electric Current â Numerical Questions
1. Magnetic Field around a Straight Current-Carrying Conductor
A current of 5 A flows through a straight conductor. The magnetic field at a point is 2 × 10âģâĩ T. Find the distance from the conductor.
āĻāĻāĻž āϏā§āĻāĻž āϤāĻžā§°āϤ ā§Ģ A āϧāĻžā§°āĻž āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšā§āĻā§āĨ¤ āĻāĻāĻž āĻŦāĻŋāύā§āĻĻā§āϤ āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§°ā§° āĻŽāĻžāύ 2 × 10âģâĩ TāĨ¤ āϤāĻžā§°āϤ āĻĒā§°āĻž āĻĻā§ā§°āϤā§āĻŦ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§:
I = 5 A
B = 2 × 10âģâĩ T
μâ = 4π × 10âģ⡠T m Aâģ¹
Formula / āϏā§āϤā§ā§°: B = (μâ I) / (2πr)
Calculation / āĻāĻŖāύāĻž:
2 × 10âģâĩ = (4π × 10âģâˇ × 5) / (2πr)
2 × 10âģâĩ = (10 × 10âģâˇ) / r
r = 0.05 m
Ans / āĻāϤā§āϤ⧰: Distance from conductor = 0.05 m = 5 cm
2. Force on a Current-Carrying Conductor
A conductor carrying 3 A current is placed in a magnetic field of 0.4 T. Length of the conductor is 0.5 m. Find the force. ā§Š A āϧāĻžā§°āĻž āĻŦāĻšāύ āĻā§°āĻž āĻāĻāĻž āϤāĻžā§° 0.4 T āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§°āϤ ā§°āĻāĻž āĻšā§āĻā§āĨ¤ āϤāĻžā§°ā§° āĻĻā§ā§°ā§āĻā§āϝ 0.5 māĨ¤ āĻŦāϞ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: B = 0.4 T, I = 3 A, L = 0.5 m
Formula / āϏā§āϤā§ā§°: F = BIL
Calculation / āĻāĻŖāύāĻž:
F = 0.4 × 3 × 0.5
F = 0.6 N
Ans / āĻāϤā§āϤ⧰: Force = 0.6 N
3. Electromagnet / Solenoid – Ampere Turns
A solenoid has 500 turns and carries a current of 2 A. Find total ampere-turns. āĻāĻāĻž āĻāϞā§āύāĻāĻĄāϤ ā§Ģā§Ļā§ĻāĻāĻž āĻĒāĻžāĻ āĻāĻā§ āĻ⧰⧠⧍ A āϧāĻžā§°āĻž āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšā§āĻā§āĨ¤ āĻāĻŽā§āĻĒāĻŋāϝāĻŧāĻžā§°-āĻĒāĻžāĻ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: N = 500, I = 2 A
Formula / āϏā§āϤā§ā§°: Ampere-turns = N × I
Calculation / āĻāĻŖāύāĻž: Ampere-turns = 500 × 2 = 1000
Ans / āĻāϤā§āϤ⧰: Ampere-turns = 1000
4. Electric Motor – Power Relation
A motor works at 12 V and draws a current of 2 A. Find power. āĻāĻāĻž āĻŦā§āĻĻā§āϝā§āϤāĻŋāĻ āĻŽā§āĻā§°ā§ 12 V āĻŦāĻŋāĻā§ąāϤ 2 A āϧāĻžā§°āĻž āϞāϝāĻŧāĨ¤ āĻā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: V = 12 V, I = 2 A
Formula / āϏā§āϤā§ā§°: P = VI
Calculation / āĻāĻŖāύāĻž: P = 12 × 2 = 24 W
Ans / āĻāϤā§āϤ⧰: Power = 24 W
5. Force Direction – Fleming’s Left Hand Rule
A conductor carrying 4 A current is placed perpendicular to a magnetic field of 0.2 T. Length of conductor is 0.25 m. Find the force. ā§Ē A āϧāĻžā§°āĻž āĻŦāĻšāύ āĻā§°āĻž āĻāĻāĻž āϤāĻžā§° 0.2 T āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§°ā§° āϏā§āϤ⧠āϞāĻŽā§āĻŦāĻāĻžā§ąā§ ā§°āĻāĻž āĻšā§āĻā§āĨ¤ āϤāĻžā§°ā§° āĻĻā§ā§°ā§āĻā§āϝ 0.25 māĨ¤ āĻŦāϞ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: B = 0.2 T, I = 4 A, L = 0.25 m
Formula / āϏā§āϤā§ā§°: F = BIL
Calculation / āĻāĻŖāύāĻž:
F = 0.2 × 4 × 0.25
F = 0.2 N
Ans / āĻāϤā§āϤ⧰: Force = 0.2 N (āĻĻāĻŋāĻļ Fleming’s Left Hand Rule āĻ āύā§āϏāĻžā§°ā§ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻž āĻšāϝāĻŧ)
Exam Tips (Magnetic Effects Chapter)
• Always write formula first
• Use SI units only
• Remember: F = BIL, B = μâI / 2πr
• Fleming’s Left Hand Rule → Direction of force
• Score = Full marks
6. Magnetic Field in a Circular Coil (Concept Question)
If the current in a circular coil is doubled, what happens to the magnetic field?
āĻāĻāĻž āĻŦā§āϤā§āϤāĻžāĻāĻžā§° āĻāĻāϞāϤ āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āϧāĻžā§°āĻž āĻĻā§āĻā§āĻŖ āĻā§°āĻŋāϞ⧠āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§°ā§° āĻāĻŋ āĻšāϝāĻŧ?
Concept / āϧāĻžā§°āĻŖāĻž: Magnetic field is directly proportional to current. āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§° āϧāĻžā§°āĻžā§° āϏā§āϤ⧠āϏā§āĻāĻžāϏā§āĻāĻŋ āϏāĻŽāĻžāύā§āĻĒāĻžāϤāĻŋāĻāĨ¤
Formula / āϏā§āϤā§ā§°: B ∝ I
Conclusion / āϏāĻŋāĻĻā§āϧāĻžāύā§āϤ: If current doubles, magnetic field also doubles.āϧāĻžā§°āĻž āĻĻā§āĻā§āĻŖ āĻš’āϞ⧠āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§°ā§ āĻĻā§āĻā§āĻŖ āĻšāϝāĻŧāĨ¤
7. Magnetic Field Produced by a Coil
A coil produces a magnetic field proportional to current. If 2 A current produces 0.02 T, find the magnetic field when current is 5 A.
āĻāĻāĻž āĻāĻāϞ⧠āĻā§āĻĒāύā§āύ āĻā§°āĻž āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§° āϧāĻžā§°āĻžā§° āϏāĻŽāĻžāύā§āĻĒāĻžāϤāĻŋāĻāĨ¤ 2 A āϧāĻžā§°āĻžāϤ 0.02 T āĻā§āώā§āϤā§ā§° āĻā§āĻĒāύā§āύ āĻšāϝāĻŧāĨ¤ 5 A āϧāĻžā§°āĻžāϤ āĻā§āώā§āϤā§ā§° āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Formula / āϏā§āϤā§ā§°: Bâ / Bâ = Iâ / Iâ
Calculation / āĻāĻŖāύāĻž:
Bâ = (0.02 × 5) / 2
Bâ = 0.05 T
Ans / āĻāϤā§āϤ⧰: Magnetic field = 0.05 T
8. Force on a Current Carrying Conductor
A conductor carrying 3 A current is placed in a magnetic field of 0.4 T. Length of conductor = 0.5 m. Find force.
3 A āϧāĻžā§°āĻž āĻŦāĻšāύ āĻā§°āĻž āĻāĻāĻž āϤāĻžā§° 0.4 T āĻā§āĻŽā§āĻŦāĻā§āϝāĻŧ āĻā§āώā§āϤā§ā§°āϤ ā§°āĻāĻž āĻšā§āĻā§āĨ¤ āϤāĻžā§°ā§° āĻĻā§ā§°ā§āĻā§āϝ = 0.5 māĨ¤ āĻŦāϞ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Formula / āϏā§āϤā§ā§°: F = BIL
Calculation / āĻāĻŖāύāĻž:
F = 0.4 × 3 × 0.5
F = 0.6 N
Ans / āĻāϤā§āϤ⧰: Force = 0.6 N
9. Electric Motor – Power Relation
A motor works at 12 V and draws 2 A current. Find power. āĻāĻāĻž āĻŽā§āĻā§°ā§ 12 V āĻŦāĻŋāĻā§ąāϤ 2 A āϧāĻžā§°āĻž āϞāϝāĻŧāĨ¤ āĻā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Formula / āϏā§āϤā§ā§°: P = VI
Calculation / āĻāĻŖāύāĻž:
P = 12 × 2
P = 24 W
Ans / āĻāϤā§āϤ⧰: Power = 24 W
10. Domestic Circuit / Motor Question
A motor operates at 12 V and draws 2 A current. Find power. āĻāĻāĻž āĻŽā§āĻā§°ā§ 12 V āϤ 2 A āϧāĻžā§°āĻž āϞāϝāĻŧāĨ¤ āĻā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Formula / āϏā§āϤā§ā§°: P = VI
Calculation / āĻāĻŖāύāĻž: P = 12 × 2 = 24 W
Ans / āĻāϤā§āϤ⧰: Power = 24 W
Exam Tips (Very Important) / āĻĒā§°ā§āĻā§āώāĻžā§° āĻāĻŋāĻĒāĻ
• Always write formula first (Sure marks)
• Remember formulas: F = BIL, B ∝ I, P = VI
• Use Right Hand Thumb Rule for magnetic field direction
• Motor & generator numericals are direct formula based
• Write units clearly (N, T, W)