Question

A square loop of side 10 cm and resistance 0.5 \(\Omega\) is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set across the plane in the northeast direction. The magnetic field decreases to zero at 0.70 s at a steady rate. Then the magnitude of the induced current during this time interval will be 

Hint:

According to {Faraday’s Law}, the induced EMF in a loop is given by the {rate of change of magnetic flux}. The induced current is found using {Ohm’s Law}, where \( I = \frac{\mathcal{E}}{R} \).

Answer 1

Step 1: Understanding Faraday’s Law of Electromagnetic Induction
According to Faraday’s Law, the induced EMF (\( \mathcal{E} \)) is given by: \[ \mathcal{E} = - \frac{d\Phi}{dt} \] where \( \Phi \) is the magnetic flux given by: \[ \Phi = B A \] Step 2: Calculating the Change in Flux
Given: \[ B_{{initial}} = 0.10 { T}, \quad B_{{final}} = 0 { T} \] \[ A = (10 { cm})^2 = (0.1 { m})^2 = 0.01 { m}^2 \] \[ dt = 0.70 { s} \] Change in flux: \[ \Delta \Phi = A (B_{{final}} - B_{{initial}}) \] \[ \Delta \Phi = (0.01) \times (0 - 0.10) \] \[ \Delta \Phi = - 10^{-3} { Wb} \] Step 3: Calculating Induced EMF
\[ \mathcal{E} = - \frac{\Delta \Phi}{dt} \] \[ \mathcal{E} = - \frac{- 10^{-3}}{0.70} \] \[ \mathcal{E} \approx 1.43 \times 10^{-3} { V} \] Step 4: Finding the Induced Current
Using Ohm’s Law, the induced current is: \[ I = \frac{\mathcal{E}}{R} \] \[ I = \frac{1.43 \times 10^{-3}}{0.5} \] \[ I = 2.86 \times 10^{-3} { A} \] Thus, the magnitude of the induced current is \( 2.86 \times 10^{-3} \) A.