Question 61:
easyA flux of \(10^{-3}\text{ Wb}\) passes through a strip having an area \(A = 0.02\text{ m}^2\). The plane of the strip is at an angle of \(60^\circ\) to the direction of a uniform field \(B\). The value of \(B\) is:
The magnetic flux through the surface is given by \(\phi = B A \sin\theta\). Substituting \(\phi = 10^{-3}\text{ Wb}\), \(A = 0.02\text{ m}^2\), and \(\theta = 60^\circ\), we get \(B = \frac{10^{-3}}{0.02 \times \sin 60^\circ} \approx 0.058\text{ T}\).
Question 62:
easyAssertion (A): The electric field created by time-varying magnetic field is non-conservative.
Reason (R): The line integral of induced electric field in a closed loop is always equal to zero.
Electric field created by time-varying magnetic field is non-conservative, which means its line integral around a closed loop is non-zero (\(\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}\)). Hence, Assertion is true but Reason is false.
Question 63:
easyA coil of \(\text{Cu}\) wire (radius \(r\), self inductance \(L\)) is bent in two concentric turns each having radius \(\frac{r}{2}\). The self-inductance now is:
Self-inductance of a coil is \(L \propto N^2 r\). When bent into \(2\) turns of radius \(r/2\), new self-inductance is \(L' \propto (2)^2 (r/2) = 2 L\).
Question 64:
easyTwo coils have mutual inductance \(0.005\text{ H}\). The current changes in the first coil according to equation \(I = I_0 \sin \omega t\), where \(I_0 = 10\text{ A}\) and \(\omega = 100\pi\text{ rad/s}\). The maximum value of EMF in the second coil is:
Induced EMF is \(e = M \frac{dI}{dt} = M I_0 \omega \cos \omega t\). Maximum EMF \(e_{max} = M I_0 \omega = 0.005 \times 10 \times 100\pi = 5\pi\text{ V}\).
Question 65:
easyA square coil of side length \(l\) is placed at centre of a large circular coil of radius \(R\), where \(R \gg l\) and coils are in same plane. The coefficient of mutual inductance of the coils is
Let a current \(I\) flow through the large circular coil, producing a magnetic field \(B = \frac{mu_0 I}{2R}\) at its center. The flux through the small square coil is \(\Phi = B'A = \left(\frac{\mu_0 I}{2R}\right) l^2\). Therefore, the mutual inductance is \(M = \frac{\Phi}{I} = \frac{\mu_0 l^2}{2R}\).
Question 66:
easyIf the flux associated with a coil varies at the rate of \(2\text{ Wb/min}\), then the induced emf in the coil is
From Faraday's law, the induced electromotive force is \(e = \frac{dPhi}{dt}\). Converting minutes to seconds: \(e = \frac{2\text{ Wb}}{60\text{ s}} = \frac{1}{30}\text{ V}\).
Question 67:
easyTwo conducting circular loops of radii \(R_1\) and \(R_2\) are placed in the same plane with their centres coinciding. If \(R_1 \gg R_2\), the mutual inductance \(M\) between them will be directly proportional to
The magnetic field produced by the larger loop 1 at the center is \(B_1 = \frac{\mu_0 I_1}{2 R_1}\). The magnetic flux through the smaller loop 2 is \(\phi_2 = B_1 A_2 = \frac{\mu_0 I_1}{2 R_1} \pi R_2^2\). Therefore, \(M = \frac{\phi_2}{I_1} = \frac{\mu_0 \pi R_2^2}{2 R_1}\), which means \(M \propto \frac{R_2^2}{R_1}\).
Question 68:
easyA step down transformer connected to an ac mains supply of \(220\text{ V}\) is made to operate at \(11\text{ V}\), \(44\text{ W}\) lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
Assuming zero power losses, the input power in the primary circuit equals the output power in the secondary circuit: \(P_{\text{in}} = P_{\text{out}} = 44\text{ W}\). Since \(P_{\text{in}} = V_p I_p\), we get \(I_p = \frac{P_{\text{in}}}{V_p} = \frac{44}{220} = 0.2\text{ A}\).
Question 69:
easyA coil has a self-inductance of 0.02 H. The current through it, is allowed to change at the rate of 4 A in \(2 \times 10^{-2}\text{ s}\). The e.m.f induced in the coil will be
Formula: \(e = L \frac{dI}{dt}\). Substituting \(L = 0.02\text{ H}\), \(dI = 4\text{ A}\) and \(dt = 2 \times 10^{-2}\text{ s}\), we get \(e = 0.02 \times 200 = 4\text{ V}\).
Question 70:
easyIn an ideal transformer, the turns ratio is \(\frac{N_p}{N_s} = \frac{1}{3}\). The ratio of \(I_s : I_p\) is equal to (symbols carry their usual meaning):
For an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s ⇒ \frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{1}{3}\).