Question

A rectangular loop of sides \(l\) and \(b\) and resistance \(R\) is kept in a region in which the magnetic field varies as \[ B=B_0\sin\omega t. \] (i) Derive expression for the emf induced in the loop. (ii) Find the effective value of current that flows in the loop.

Hint:

For a loop placed in a magnetic field varying as \[ B=B_0\sin\omega t, \] first calculate the magnetic flux using \[ \Phi_B=BA. \] Then apply Faraday's law \[ e=-\frac{d\Phi_B}{dt}. \] For a sinusoidal current, \[ I_{\text{rms}}=\frac{I_0}{\sqrt{2}}. \] Remember that the peak induced emf is \[ e_0=B_0A\omega. \]

Answer 1

Concept: Whenever the magnetic flux linked with a closed conducting loop changes with time, an emf is induced in the loop. This phenomenon is known as electromagnetic induction. According to Faraday's law of electromagnetic induction, \[ e=-\frac{d\Phi_B}{dt}, \] where \[ \Phi_B \] is the magnetic flux linked with the loop. The negative sign represents Lenz's law, which states that the induced emf always opposes the cause producing it. In the present problem, the magnetic field varies sinusoidally with time. Therefore, the magnetic flux through the loop also varies sinusoidally, producing an alternating emf and hence an alternating current in the circuit.

Step 1: Determine the area of the rectangular loop. The rectangular loop has sides \[ l \] and \[ b. \] Hence, the area enclosed by the loop is \[ A=lb. \]

Step 2: Write the expression for magnetic flux through the loop. The magnetic field is given as \[ B=B_0\sin\omega t. \] Assuming that the magnetic field is perpendicular to the plane of the loop, the magnetic flux linked with the loop is \[ \Phi_B=BA. \] Substituting \[ A=lb, \] we obtain \[ \Phi_B=(B_0\sin\omega t)(lb). \] Therefore, \[ \boxed{\Phi_B=B_0lb\sin\omega t}. \] This expression shows that the magnetic flux varies sinusoidally with time.

Step 3: Apply Faraday's law to obtain the induced emf. According to Faraday's law, \[ e=-\frac{d\Phi_B}{dt}. \] Substituting the expression for flux, \[ e = -\frac{d}{dt} \left( B_0lb\sin\omega t \right). \] Since \(B_0\), \(l\) and \(b\) are constants, \[ e = -B_0lb \frac{d}{dt} (\sin\omega t). \] Using \[ \frac{d}{dt}(\sin\omega t) = \omega\cos\omega t, \] we get \[ e=-B_0lb\omega\cos\omega t. \] Hence, the induced emf is \[ \boxed{e=-B_0lb\omega\cos\omega t}. \] The negative sign indicates the direction of induced emf according to Lenz's law.

Step 4: Determine the maximum value of induced emf. Comparing \[ e=-B_0lb\omega\cos\omega t \] with the standard form \[ e=e_0\cos\omega t, \] we find that the maximum or peak value of induced emf is \[ \boxed{e_0=B_0lb\omega}. \]

Step 5: Obtain the expression for induced current. Using Ohm's law, \[ i=\frac{e}{R}. \] Substituting the value of induced emf, \[ i = \frac{-B_0lb\omega\cos\omega t}{R}. \] Therefore, \[ \boxed{i=-\frac{B_0lb\omega}{R}\cos\omega t}. \] This current is alternating in nature because it varies periodically with time.

Step 6: Determine the peak value of current. Comparing the current expression with \[ i=i_0\cos\omega t, \] we obtain \[ \boxed{i_0=\frac{B_0lb\omega}{R}}. \] This is the maximum value of induced current flowing through the loop.

Step 7: Calculate the effective (rms) value of current. For a sinusoidally varying alternating current, \[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}}. \] Substituting the value of \(I_0\), \[ I_{\text{rms}} = \frac{1}{\sqrt{2}} \left( \frac{B_0lb\omega}{R} \right). \] Hence, \[ \boxed{ I_{\text{rms}} = \frac{B_0lb\omega}{\sqrt{2}\,R} }. \] This is the effective value of current flowing in the loop. Final Answers: \[ \boxed{ e=-B_0lb\omega\cos\omega t } \] and \[ \boxed{ I_{\text{rms}} = \frac{B_0lb\omega}{\sqrt{2}\,R} }. \]