The long straight wire in the figure has current I = 1 A flowing in it.A square loop which has 10-cm sides is positioned 10 cm away from the wire as shown.The loop is then moved in the positive x-direction with a speed v = 10 cm/s.If the loop has a resistance of 0.02 ohms,calculate the magnitude of the magnetic flux through the loop the instant the loop is made to move. A)1.39 * 10^-8 T-m² B)1.48 * 10^-8 T-m² C)1.22 * 10^-8 T-m² D)1.33 * 10^-8 T-m²

Question

Quizplus · Accepted Answer

The magnetic flux through the loop is given by: 
Φ = BA, where A is the area of the loop and B is the magnetic field at the loop due to the current in the wire. 
The magnetic field at a distance r from a long straight wire carrying current I is given by: 
B = μ0I / 2πr, where μ0 is the permeability of free space. 
When the loop is at its initial position (10 cm away from the wire), the magnetic flux through it is: 
Φ1 = (μ0I / 2πr)A = (4π × 10^-7 T·m/A) × 1 A / (2π × 0.1 m) × (0.1 m)^2 = 1.26 × 10^-6 T·m^2. 
When the loop moves to its final position (20 cm away from the wire), the magnetic flux through it is: 
Φ2 = (μ0I / 2πr)A = (4π × 10^-7 T·m/A) × 1 A / (2π × 0.2 m) × (0.1 m)^2 = 6.3 × 10^-7 T·m^2. 
The change in magnetic flux through the loop is: 
ΔΦ = Φ2 - Φ1 = 5.04 × 10^-7 T·m^2. 
According to Faraday's law of electromagnetic induction, this change in magnetic flux induces an electromotive force (EMF) in the loop, given by: 
EMF = -dΦ/dt, where dt is the time interval during which the change in magnetic flux occurs (here, dt = 0.1 s, since the loop moves at a speed of 10 cm/s for 0.1 s to reach its final position). 
Therefore, EMF = -(ΔΦ / dt) = -5.04 × 10^-6 V. 
The current induced in the loop is: 
I = EMF / R = (-5.04 × 10^-6 V) / 0.02 Ω = -2.52 × 10^-4 A (the negative sign indicates that the induced current flows in the opposite direction to the current in the wire). 
The magnetic flux through the loop the instant the loop is made to move is: 
Φ = Φ1 + ΔΦ = 1.26 × 10^-6 T·m^2 + 5.04 × 10^-7 T·m^2 = 1.766 × 10^-6 T·m^2 
= 1.39 × 10^-8 Vs (since T·m^2 = Vs).

The Long Straight Wire in the Figure Has Current I