A magnet oscillates in Earth's horizontal magnetic field $B_H$ with time period T. An external horizontal field of magnitude 3 $B_H/4$ is then applied opposite to Earth's horizontal 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. An external horizontal field of magnitude 3 $B_H/4$ is then applied opposite to Earth's horizontal field.

The new time period is:

Options

G1 Physics · Accepted Answer

The opposing field reduces the net field to B n e t B_net B n ​ e t = B H B_H B H ​ - 3 B H / 4 B_H/4 B H ​ /4 = B H / 4. B_H/4. B H ​ /4. For small oscillations, T ∝ 1/√B. Hence, T'/T = √( B H / ( B H / 4 ) ) B_H/(B_H/4)) B H ​ / ( B H ​ /4 )) = √4 = 2. Therefore, T' = 2T.

A magnet oscillates in Earth's horizontal magnetic field B H B_H B H ​ with time period T. An external horizontal field of magnitude 3 B H / 4 B_H/4 B H ​ /4 is then applied opposite to Earth's horizontal 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. An external horizontal field of magnitude 3 B H / 4 B_H/4 B H /4 is then applied opposite to Earth's horizontal field. The new time period is: