Question

A short magnet oscillates in vibration magnetometer with a frequency 10 Hz;
where horizontal component of earth's magnetic field is 12 ut. A downward
current of 15A is established in a long vertical wire placed 20 cm west of the
magnet. New frequency is
A) 2.5 Hz
B) 5 Hz
C) 9 Hz
D) 4 Hz​

Answer 1

Answer:

Explanation:

4Hz

2.5Hz

9Hz

15Hz

Answer :

D

Solution :

Frequency, v=12πMH−−−−√I

v=12πM(B+H)−−−−−−−−−√I

where, B=magnetic field due to downward conductor

Frequency, v=12πMH−−−−√I

v=12πM(B+H)−−−−−−−−−√I

where, B=magnetic field due to downward conductor

Then B=μ04π.2Ia

B=10−7×2×1520×10−2

B=15μT

v'v=B+BHB−−−−−−−√