Question

A uniform magnetic field$\overrightarrow{\mathbf{B}}$is perpendicular to the plane of a circular loop of diameter 10 cmformed from wire of diameter2.5 mm and resistivity$\mathbf{1}\mathbf{.}\mathbf{69}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{\Omega }\mathbf{-}\mathit{m}$. At what rate must the magnitude of$\overrightarrow{\mathbf{B}}$change to induce a 10Acurrent in the loop?

Short answer

The rate of change of the magnetic field is$1.4\frac{T}{s}$ .

Step-by-step solution

Given

i) Diameter of the loop$d_L=10cm=0.10m$

ii) Diameter of the wire$d_w=2.5mm=00.0025m$

iii) Force acting on the particle $\rho =1.69\times {10}^{-8}\Omega -m$

iv) Induced current i = 10 A$i=10A$

Determining the concept

By using expression for Magnetic flux, Faraday’s law, Resistivity, Ohm’s law, find the rate of change of the magnetic field.

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

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:

i) Magnetic flux,$\phi =\oint BdA$

ii) Faraday’s law,$\epsilon =-\frac{d\phi }{dt}$

iii) Resistance,$d=\frac{\rho L}{A}$

iv) Ohm’s law ,$V=IR$

Where,$\Phi$is magnetic flux, B is magnetic field, A is area, I is current, R is resistance,𝜀 is emf, L is length, V IS voltage,𝜌 is resistivity.

Determining the rate of change of the magnetic field

Magnetic flux through the loop area is given by,

$$\begin{gathered} \phi =\oint Bd{A}_L \\ \phi =B{A}_L \end{gathered}$$

Induced emf loop in the loop is,

$$\begin{gathered} \epsilon =\frac{d\phi }{dt} \\ \epsilon =\frac{d\left(B{A}_L\right)}{dt} \\ \frac{\epsilon }{A_L}=\frac{dB}{dt}........................................................\left(1\right) \end{gathered}$$

Area of the loop $A_L$is given by,

$A_L=\frac{\tau \tau d_L^2}{4}.............................................................\left(2\right)$

By Ohm’s law,

$\epsilon =IR................................................................\left(3\right)$

To find the resistance of the wire,

$R=\frac{\rho L}{A_W}$

Length of the wire L= circumference of the loop =$\tau \tau d_L$

Area of the cross section of the wire is given by,$A_W=\frac{\tau \tau d_W^2}{4}$

Therefore, the resistance of the wire is,

$$\begin{gathered} R=\rho \frac{\tau \tau d_L}{{\displaystyle \frac{\tau \tau d_W^2}{4}}} \\ R=\frac{4\rho d_L}{d_W^2} \end{gathered}$$

Using in equation 3,

$$\begin{gathered} \epsilon =IR \\ \epsilon =I\times \frac{4\rho d_L}{d_W^2}.........................................................\left(4\right) \end{gathered}$$

Using equations 2 and 4 in 1,

$$\begin{gathered} \frac{\epsilon }{A_L}=\frac{dB}{dt} \\ \frac{dB}{dt}=I\times \frac{4\rho d_L}{d_W^2}\times \frac{4}{\tau \tau d_L^2} \\ \frac{dB}{dt}=\frac{16Ip}{\tau \tau d_L{d}_W^2} \end{gathered}$$

Using all the given values,

$$\begin{gathered} \frac{dB}{dt}=\frac{16\times 10\times 1.69\times {10}^{-8}}{3.14\times 0.10\times 0.{0025}^2} \\ \frac{dB}{dt}=1.37T/s \\ \frac{dB}{dt}=1.4T/s \end{gathered}$$

Hence, the rate of change of the magnetic field is,$\frac{dB}{dt}=1.4\frac{T}{s}$ .

Therefore, by using expression for Magnetic flux, Faraday’s law, Resistivity, Ohm’s law, the rate of change of the magnetic field can be found.