Problem
A long solenoid with 15 turns per cm has a small loop of area 2.0cm2 placed inside the solenoid, normal to its axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced e.m.f. in the loop while the current is changing?
Data
- Number of turns per cm: 15turns/cm=1500turns/m
- Turns per unit length: n=1500turns/m
- Area of the loop:
A=2.0cm2=2.0×(10−2)2m2=2.0×10−4m2
- Initial current: Ii​=2.0A
- Final current: If​=4.0A
- Time interval: Δt=0.1s
To find:
- Induced e.m.f. ε.
Prerequisite Concepts
- Magnetic field inside a solenoid:
B=μ0​nI
where:
- μ0​=4π×10−7T⋅m/A is the permeability of free space,
- n is the number of turns per unit length,
- I is the current.
- Change in magnetic field:
ΔB=μ0​nΔI
- Faraday’s law of electromagnetic induction:
ε=ΔtΔΦ​
where:
ΔΦ=ΔB⋅A
Solution
Step 1: Calculate the Change in Magnetic Field (ΔB)
- Use the formula:
ΔB=μ0​nΔI
- Substitute the values:
ΔB=(4π×10−7)⋅1500⋅(4−2)
- Simplify:
ΔB=4×3.14×10−7⋅1500⋅2
ΔB=3.77×10−4T.
Step 2: Calculate the Induced e.m.f. (ε)
- Use Faraday’s law:
ε=ΔtΔΦ​=ΔtΔB⋅A​
- Substitute the values:
ε=0.1(3.77×10−4)⋅(2.0×10−4)​
- Simplify:
ε=0.17.54×10−8​=7.54×10−5V.
Answer
The induced e.m.f. in the loop is:
ε=7.54×10−5V.