Question

A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The square loop has a resistance of 30 W. The solenoid has 500 turns and is 0.3 m long. The current in the solenoid is increasing at a constant rate of 0.7 A/s. What is the current flowing in the square loop?

Answer 1

Answer:

Number of turns per metre, n= 500/0.3= (5000/3)m^-1

Cross sectional areaof the square loop of wire, A= (0.1^2)m^2= 0.01m^2dB/dt= μn(dI/dt)= (4.00π x10^-7)(5000/3)(0.7)= 1.46608x10^-3T/s

The induced emf in the square loop of wire, ε= the rate of change of magnetic flux of the square loop of wire(dΦ/dt)= A(dB/dt)= (0.01)(1.46608x10^-3)= 0.0146608x10^-5VA

current flows in the square loop of wire since a potential difference(induced emf in this case) exists. Its magnitude,

I= ε/R where R is the resistance of the square loop of wire.

I= (0.0146608x10^-5)/30= 4.89x10^-7A

Answer 2

Answer:

$I=9.6*10^{-8}A$

Explanation:

We first calculate the change in magnetic flux ($\Phi_B$) due to the change in the magnitude of magnetic field (B), since que current (I) in the solenoid changes. B and $\Phi_B$ are given by:

$\Phi_B=ABcos\theta\\\frac{\Delta \Phi_B}{\Delta t}=A\frac{\Delta B}{\Delta t}cos\theta(1)\\B=n\mu_0I\\\frac{\Delta B}{\Delta t}=N\mu_0\frac{\Delta I}{\Delta t}(2)$

In this case, the magnetic field is always perpendicular to the square loop. We replace (2) in (1):

$\frac{\Delta \Phi_B}{\Delta t}=An\mu_0\frac{\Delta I}{\Delta t}cos(90^\circ)$

According to Faraday's law:

$\epsilon=-\frac{\Delta \Phi_B}{\Delta t}\\\epsilon=-An\mu_0\frac{\Delta I}{\Delta t}$

The minus sign means that the resulting current will have a direction such that the induced magnetic field will oppose the original change in flux. Here, n is the number of turns per unit length and A is the solenoid area

$\epsilon=(\pi r^2)(\frac{N}{L})\mu_0\frac{\Delta I}{\Delta t}\\\epsilon=\pi(0.025m)^2(\frac{500}{0.3m})(4\pi*10^{-7}\frac{T*m}{A})(0.7\frac{A}{s})\\\epsilon=2.88*10^{-6}V$

Recall that $\epsilon$ is an induced voltage. We use the Ohm's law to calculate the current:

$V=IR\\I=\frac{V}{R}\\I=\frac{2.88*10^{-6}V}{30\Omega}\\I=9.6*10^{-8}A$