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
B
0.2 s
C
0.26 s
D
0.16 s

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  $$\displaystyle T_{new}=2(0.1)=0.2s$$PhysicsNCERTStandard XII

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More

People also searched for
View More