Question

Question: A small circular loop of area 2.00$\mathit{c}{\mathit{m}}^{\mathbf{2}}$200 Ais placed in the plane of, and concentric with, a large circular loop of radius 1.00 t = 0.500 s. The current in the large loop is changed at a constant rate from -200 A to(a change in direction) in a time of 1.00 s, starting at t = 0. (a) What is the magnitude of the magnetic field Bat the center of the small loop due to the current in the large loop at t = 0?(b) What is the magnitude of the magnetic field $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$at the center of the small loop due to the current in the large loop at t = 0.500 s?(c) What is the magnitude of the magnetic fieldat the center of the small loop due to the current in the large loop at t =1.00s? (d) Fromt = 0 to t = 1.00 s, is$\stackrel{\rightharpoonup }{\mathit{B}}$. reversed? Because the inner loop is small, assume$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$is uniform over its area. (e) What emf is induced in the small loop at ?

Short answer

The magnitude of the magnetic field at the center of the small loop due to the current in the large loop at

1. $t=0s\mathrm{is}\mathrm{B}=1.26\times {10}^{-4}\mathrm{T}$
2. $t=0.500s\mathrm{is}\mathrm{B}=0.0\mathrm{T}$
   
3. $t=1.00s\mathrm{is}\mathrm{B}=-1.26\times {10}^{-4}\mathrm{T}$
   
4. Yes, the magnetic field is reversed.
5. The emf induced in the small loop at $t=0.500s\mathrm{is}\epsilon =5.04\times {10}^{-8}\mathrm{V}$.

Step-by-step solution

Step 1: Given

1. Small loop area$a=2.00c{m}^2=2.00\times {10}^{-4}{\mathrm{m}}^2$
2. Large circular loop radius $R=1.00\mathrm{m}$
3. Current in large loop changed at constant rate from 200A to -200A in time t = 1.00s

Determining the concept

The current is constantly changing with time. Determine the current at different time instances. For the circular loop, the magnetic field depends on the current and radius. The emf is related to the change in flux with time.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

$$\begin{gathered} B=\frac{{\mu }_{0}l}{2R} \\ \epsilon =-\frac{d\mathbf{\varphi }}{dt} \end{gathered}$$

Where,$\mathbf{\varphi }$is magnetic flux, B is magnetic field, I is current, R is resistance, 𝜀 is emf, , 𝜇₀is permeability.

(a) Determining the magnitude of the magnetic field at t = 0s

Current in the large loop changed at constant rate from 200A to -200A in time

t = 1.00s

$l=l_0\left(1-2t\right)$and $l_0=200A$
At role="math" localid="1661852445388" $t=0\mathrm{s},l=200\mathrm{A}$

At $t=0.500\mathrm{s},l=0.00\mathrm{A}$

At $t=1.00\mathrm{s},l=-200\mathrm{A}$

Hence,

At t = 0s,

$$\begin{gathered} B=\frac{{\mu }_{0}l}{2R}=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)200\mathrm{A}}{2\left(1.00m\right)} \\ B=1.26\times {10}^{-4}T \end{gathered}$$

Hence, the magnitude of the magnetic field at the center of the small loop due to the current in the large loop at t = 0S is $B=1.26\times {10}^{-4}T$.

(b) Determining the magnitude of the magnetic field at t = 0.500s

At t = 0.500s,

B = 0.0T

Hence, the magnitude of the magnetic field at the center of the small loop due to the current in the large loop at t = 0.500s is B = 0.0T

(c) Determining the magnitude of the magnetic field at

At t = 1.00s,

$$\begin{gathered} B=\frac{{\mu }_{0}l}{2R} \\ B=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)\left(-200A\right)}{2\left(100m\right)} \\ B=-1.26\times {10}^{-4}T \end{gathered}$$

Hence, the magnitude of the magnetic field at the center of the small loop due to the current in the large loop at$t=1.00s$ is$B=-1.26\times {10}^{-4}T$

(d) Determining is magnetic field reversed

The current changed from 200A to -200A the magnetic field has reversed from $1.26\times {10}^{-4}$ T to$-1.26\times {10}^{-4}$T in time t = 0s to t = 1.00s.

Hence, the magnetic field is reversed.

(e) Determining the emf induced in the small loop at t = 0.500s

The emf induced in a small loop is given by,

$\epsilon =\frac{d\mathbf{\varphi }}{dt}$

Also, the flux can be written in terms of current and loop area,

$\mathrm{\varphi }=\frac{{\mathrm{\mu }}_0\mathrm{lA}}{2\mathrm{R}}$

Where, A is the area of the small loop i.e.a = 2.00$\times$${10}^{-4}{\mathrm{m}}^2$

role="math" localid="1661853987331" $$\begin{gathered} \epsilon =-\frac{d\mathbf{\varphi }}{dt} \\ \epsilon =-\frac{d}{dt}\frac{{\mathrm{\mu }}_0\mathrm{lA}}{2\mathrm{R}} \\ \epsilon =-\frac{{\mathrm{\mu }}_0\mathrm{lA}}{2\mathrm{R}}\frac{dl}{dt} \\ \epsilon =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)2.00\times {10}^{-4}{\mathrm{m}}^2\left(-200\mathrm{A}-\left(200\mathrm{A}\right)\right)}{2\left(1.00m\right)\left(1.00s\right)} \\ \epsilon =5.04\times {10}^{-8}V \end{gathered}$$

Hence, the emf induced in the small loop at t = 0.500s is $\epsilon =5.04\times {10}^{-8}\mathrm{V}$.

Therefore, find the magnetic field for the circular loop at different time instances using the current at the given time and radius. The emf is determined with the change in flux.