question_answer

A short magnet oscillating in vibration magnetometer with a frequency 10 Hz. A downward current of 15 A is established in a kmg vertical wire placed 20 cm to the West of the magnet. The new frequency of the short magnet is (the horizontal component of earths magnetic field is 12 \[\mu \]T)

A)                  4 Hz                                      

B)                  2.5 Hz

C)                  9 Hz                                      

D)                  15 Hz

Correct Answer: D

Solution :

                 \[v=\frac{1}{2\pi }\sqrt{\frac{M{{B}_{H}}}{I}}\]                 \[v=\frac{1}{2\pi }\sqrt{\frac{M(B+{{B}_{H}})}{I}}\]where B = Magnetic field due to downward conductor.                                 Then,    \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i}{a}\]                                 \[B={{10}^{-7}}\times \frac{2\times 15}{20\times {{10}^{-2}}}\]                                 \[B=15\mu T\]                 \[\therefore \]  \[\frac{v}{v}=\sqrt{\frac{B+{{B}_{H}}}{B}}\]                                 \[\frac{v}{10}=\sqrt{\frac{15+12}{12}}=1.5\]                                 \[v=15\,Hz\]