Question

A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth's horizontal magnetic field is 24 μT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.

Answer 1

The new time period $T_2$  is 0.076 s

Explanation:

Step 1:

Given That  

In this question magnet is oscillating in an oscillation magnetometer.

Time Period $T_1$ = 0.10 s        

Horizontal Magnetic field $\overrightarrow{B H}=24 \mu T=24 \times 10^{6} T$

Current I = 18A

d= 20 cm = 0.2 m

Here it is required to identify the net magnetic field.

Step 2:

The formula for net magnetic field is  

$\vec{B}=\overrightarrow{B H} \cdot \vec{B}(\text { wires })$

$\vec{B}=\overrightarrow{B H}-\frac{\mu o l}{2 \pi r}$

Here keep the values in the place of formula

The value of $\mu o=4 \pi X 10^{-7}$

$\vec{B}=24 \times 10^{-6}-\frac{4 \pi X 10^{-7} X 18}{2 \pi X 0.2}$

$\vec{B}=24 \times 10^{-6}-\frac{2 \times 10^{-7} X 18}{0.2}$

$\vec{B}=14 \times 10^{-6}$

$\mathrm{T}=2 \pi \sqrt{\frac{I}{M B H}}$

Here we will get

$\frac{T 1}{T 2}=\sqrt{\frac{B}{B H}}$

$\frac{0.1}{T 2}=\sqrt{\frac{14 \times 10^{-6}}{24 \times 10^{-6}}}$

Thus $T_2$  Value is

$T 2=\sqrt{\frac{0.01 \times 14}{24}}$

After calculating , T2 value is

$T_2$ = 0.076s

Answer 2

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The new timeperiod T2 is 0.076 sec.