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Standard XII

Physics

Intro to Simple Pendulum

A short magne...

Question

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

0.1 s

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0.2 s

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0.26 s

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0.16 s

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Solution

The correct option is B $0.2 s$
$B = \frac{μ_{0} I}{2 π d} = \frac{4 π \times 10^{- 7} \times 18}{2 π \times 0.2}$
$= 18 μ T$ South or North
$\frac{T_{n e w}}{T} = \sqrt{\frac{B_{H}}{B_{H} - B}} = \sqrt{\frac{24}{24 - 18}} = 2$
Thus $T_{n e w} = 2 (0.1) = 0.2 s$

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A short magnet is executing oscillation in Earth's horizontal magnetic field of 24 μT. 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 :

The correct option is B 0.2 sB=μ0I2πd=4π×10−7×182π×0.2=18μT South or NorthTnewT=√BHBH−B=√2424−18=2Thus Tnew=2(0.1)=0.2s

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

The correct option is B $0.2 s$
$B = \frac{μ_{0} I}{2 π d} = \frac{4 π \times 10^{- 7} \times 18}{2 π \times 0.2}$
$= 18 μ T$ South or North
$\frac{T_{n e w}}{T} = \sqrt{\frac{B_{H}}{B_{H} - B}} = \sqrt{\frac{24}{24 - 18}} = 2$
Thus $T_{n e w} = 2 (0.1) = 0.2 s$