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

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

`I=9.6*10^(-8)A`

Step-by-step 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\\(\Delta \Phi_B)/(\Delta t)=A(\Delta B)/(\Delta t)cos\theta(1)\\B=n\mu_0I\\(\Delta B)/(\Delta t)=N\mu_0(\Delta I)/(\Delta t)(2)`

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

`(\Delta \Phi_B)/(\Delta t)=An\mu_0(\Delta I)/(\Delta t)cos(90^\circ)`

According to Faraday's law:

`\epsilon=-(\Delta \Phi_B)/(\Delta t)\\\epsilon=-An\mu_0(\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)((N)/(L))\mu_0(\Delta I)/(\Delta t)\\\epsilon=\pi(0.025m)^2((500)/(0.3m))(4\pi*10^(-7)(T*m)/(A))(0.7(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=(V)/(R)\\I=(2.88*10^(-6)V)/(30\Omega)\\I=9.6*10^(-8)A`