Skip to content

4 - N u m e r i c a l   7

Problem

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm22.0 \, \mathrm{cm}^2 placed inside the solenoid, normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A2.0 \, \mathrm{A} to 4.0 A4.0 \, \mathrm{A} in 0.1 s0.1 \, \mathrm{s}, what is the induced e.m.f. in the loop while the current is changing?

Data

  • Number of turns per cm: 15 turns/cm=1500 turns/m15 \, \mathrm{turns/cm} = 1500 \, \mathrm{turns/m}
  • Turns per unit length: n=1500 turns/mn = 1500 \, \mathrm{turns/m}
  • Area of the loop:
A=2.0 cm2=2.0×(10−2)2 m2=2.0×10−4 m2 A = 2.0 \, \mathrm{cm}^2 = 2.0 \times (10^{-2})^2 \, \mathrm{m}^2 = 2.0 \times 10^{-4} \, \mathrm{m}^2
  • Initial current: Ii=2.0 AI_i = 2.0 \, \mathrm{A}
  • Final current: If=4.0 AI_f = 4.0 \, \mathrm{A}
  • Time interval: Δt=0.1 s\Delta t = 0.1 \, \mathrm{s}

To find:

  • Induced e.m.f. ε\varepsilon.

Prerequisite Concepts

  1. Magnetic field inside a solenoid:
B=μ0nI B = \mu_0 n I

where:

  • μ0=4π×10−7 Tâ‹…m/A\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T \cdot m/A} is the permeability of free space,
  • nn is the number of turns per unit length,
  • II is the current.
  1. Change in magnetic field:
ΔB=μ0nΔI \Delta B = \mu_0 n \Delta I
  1. Faraday’s law of electromagnetic induction:
ε=ΔΦΔt \varepsilon = \frac{\Delta \Phi}{\Delta t}

where:

ΔΦ=ΔB⋅A \Delta \Phi = \Delta B \cdot A

Solution

Step 1: Calculate the Change in Magnetic Field (ΔB\Delta B)

  1. Use the formula:
ΔB=μ0nΔI \Delta B = \mu_0 n \Delta I
  1. Substitute the values:
ΔB=(4π×10−7)⋅1500⋅(4−2) \Delta B = (4\pi \times 10^{-7}) \cdot 1500 \cdot (4 - 2)
  1. Simplify:
ΔB=4×3.14×10−7⋅1500⋅2 \Delta B = 4 \times 3.14 \times 10^{-7} \cdot 1500 \cdot 2 ΔB=3.77×10−4 T. \Delta B = 3.77 \times 10^{-4} \, \mathrm{T}.

Step 2: Calculate the Induced e.m.f. (ε\varepsilon)

  1. Use Faraday’s law:
ε=ΔΦΔt=ΔB⋅AΔt \varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{\Delta B \cdot A}{\Delta t}
  1. Substitute the values:
ε=(3.77×10−4)⋅(2.0×10−4)0.1 \varepsilon = \frac{(3.77 \times 10^{-4}) \cdot (2.0 \times 10^{-4})}{0.1}
  1. Simplify:
ε=7.54×10−80.1=7.54×10−5 V. \varepsilon = \frac{7.54 \times 10^{-8}}{0.1} = 7.54 \times 10^{-5} \, \mathrm{V}.

Answer

The induced e.m.f. in the loop is:

ε=7.54×10−5 V.\varepsilon = 7.54 \times 10^{-5} \, \mathrm{V}.