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Question

A short magnet is executing oscillation in Earth's horizontal magnetic field of $$24\ \mu T$$. Time for $$20$$ vibrations is $$2\ s$$ . An upward electric current of $$18\ A$$ is established in a vertical wire placed $$20 \ cm$$  east of the magnet. The new time period is :


A
0.1 s
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B
0.2 s
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C
0.26 s
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D
0.16 s
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Solution

The correct option is B $$0.2\ s$$
$$\displaystyle B=\dfrac{\mu_0 I}{2\pi d}=\frac{4\pi\times10^{-7}\times18}{2\pi\times 0.2}$$
$$\displaystyle =18\mu T$$ South or North
$$\displaystyle\dfrac{T_{new}}{T}=\sqrt{\dfrac{B_H}{B_H-B}}=\sqrt\frac{24}{24-18}=2$$
Thus  $$\displaystyle  T_{new}=2(0.1)=0.2s$$

Physics
NCERT
Standard XII

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