The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr^{3+ ion (Atomic no. : Cr = 24) is:
3.57 B.M.
Step 1: Understanding the magnetic moment. The magnetic moment depends on the number of unpaired electrons. For Cr^{3+} (with atomic number 24), the electron configuration is \( [Ar]3d^3 \), meaning there are 3 unpaired electrons.
Step 2: Calculation. The magnetic moment (\(\mu\)) is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] Where \( n \) is the number of unpaired electrons. For Cr^{3+} (3 unpaired electrons), we get: \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \]
Step 3: Conclusion. Thus, the magnetic moment is 3.87 B.M., corresponding to option (B). \vspace{10pt}
A circular coil of diameter 15 mm having 300 turns is placed in a magnetic field of 30 mT such that the plane of the coil is perpendicular to the direction of the magnetic field. The magnetic field is reduced uniformly to zero in 20 ms and again increased uniformly to 30 mT in 40 ms. If the EMFs induced in the two time intervals are \( e_1 \) and \( e_2 \) respectively, then the value of \( e_1 / e_2 \) is: