If a current is passed through a spring then the spring will:
Current flows in the same direction in adjacent turns of the spring. Since parallel currents in the same direction attract each other, the turns of the spring are pulled closer, causing the spring to compress.
In a step up transformer, the voltage in the primary is \(220\text{ V}\) and the current is \(5\text{ A}\). The secondary voltage is found to be \(22000\text{ V}\). The current in the secondary coil (neglect losses) is:
In an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s\). Therefore, \(I_s = \frac{V_p I_p}{V_s} = \frac{220 \times 5}{22000} = 0.05\text{ A}\).
An aeroplane in which the distance between the tips of the wings is \(50\text{ m}\) is flying horizontally with a speed of \(360\text{ km/hr}\) over a place where the vertical component of earth’s magnetic field is \(2 \times 10^{-4}\text{ Wbm}^{-2}\). The potential difference between the tips of the wings would be:
Induced EMF is \(e = B_v l v\). Converting speed: \(v = 360\text{ km/h} = 100\text{ m/s}\). Thus, \(e = (2 \times 10^{-4}) \times 50 \times 100 = 1.0\text{ V}\).
A current of \(2\text{ A}\) is increasing at a rate of \(4\text{ A/s}\) through a coil of inductance \(2\text{ H}\). The energy stored in the inductor per unit time in given instant is:
Formula for rate of change of energy in an inductor is \(\frac{dU}{dt} = LI\frac{dI}{dt}\). Given \(L = 2\text{ H}\), \(I = 2\text{ A}\), and \(\frac{dI}{dt} = 4\text{ A/s}\), we get \(\frac{dU}{dt} = 2 \times 2 \times 4 = 16\text{ J/s}\).
Two coils of self inductance \(2\text{ mH}\) and \(8\text{ mH}\) are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:
For complete coupling, the coupling coefficient \(k = 1\). The mutual inductance is given by \(M = k\sqrt{L_1 L_2} = \sqrt{2 \times 8} = 4\text{ mH}\).
A 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 \(B = \frac{10^{-3}}{0.02 \times \sin 60^\circ} \approx 0.058\text{ T}\).
Assertion (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.
A 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 \(L' \propto (2)^2 (r/2) = 2 L\).
Two 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 \(e_{max} = M I_0 \omega = 0.005 \times 10 \times 100\pi = 5\pi\text{ V}\).
A 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 \(M = \frac{\Phi}{I} = \frac{\mu_0 l^2}{2R}\).