
Step 1: Magnetic Field Due to the Arc
For the arc with radius \( a \) and angle \( \frac{3\pi}{2} \), the magnetic field at the origin is: \[ B_1 = \frac{\mu_0 I}{4\pi a} \]
Step 2: Magnetic Field Due to the Straight Segment
For the straight segment of the wire: \[ B_2 = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]
Step 3: Magnetic Field Due to Other Segments
Since the magnetic field due to the straight segments at the origin is zero: \[ B_3 = 0 \]
Step 4: Calculate the Total Magnetic Field
Thus, the total magnetic field at the origin is: \[ B = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]