Magnetism Question 69
Question: A short magnet of moment 6.75 Am2 produces a neutral point on its axis. If horizontal component of earth’s magnetic field is $ 5\times {{10}^{-5}}Wb/m^{2} $ , then the distance of the neutral point should be
[SCRA 1994]
Options:
A) 10 cm
B) 20 cm
C) 30 cm
D) 40 cm
Show Answer
Answer:
Correct Answer: C
Solution:
At neutral point $ | \begin{matrix} ,\text{Magnetic field due } \ \text{to magnet} \ \end{matrix} |=,| \begin{matrix} \text{Magnetic field due } \ \text{to earth} \ \end{matrix} | $
$ \frac{{\mu _{0}}}{4\pi }.\frac{2M}{d^{3}}=5\times {{10}^{-5}}\Rightarrow {{10}^{-7}}\times \frac{2\times 6.75}{d^{3}}=5\times {{10}^{-5}} $
$ \Rightarrow d=0.3\ m=30\ cm $
BETA
