A magnet oscillates in Earth's horizontal magnetic field $B_H$ with time period T. At the same place, a uniform horizontal external magnetic field of magnitude 3 $B_H$ is applied parallel to Earth's field.
The new time period is:
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Question

A magnet oscillates in Earth's horizontal magnetic field $B_H$ with time period T. At the same place, a uniform horizontal external magnetic field of magnitude 3 $B_H$ is applied parallel to Earth's field.

The new time period is:

Options

G1 Physics · Accepted Answer

Initially, T = 2π√(I/M B H ) . B_H). B H ​ ) . The external field is parallel to Earth's field, so B' = B H B_H B H ​ + 3 B H B_H B H ​ = 4 B H . B_H. B H ​ . Therefore, T' = 2π√(I/4M B H ) B_H) B H ​ ) = T/2.

A magnet oscillates in Earth's horizontal magnetic field B H B_H B H ​ with time period T. At the same place, a uniform horizontal external magnetic field of magnitude 3 B H B_H B H ​ is applied parallel to Earth's field. The new time period is:

A magnet oscillates in Earth's horizontal magnetic field B H B_H B H with time period T. At the same place, a uniform horizontal external magnetic field of magnitude 3 B H B_H B H is applied parallel to Earth's field. The new time period is: