Magnetism Question 136
Question: The radius of the coil of a Tangent galvanometer. which has 10 turns is 0.1m. The current required to produce a deflection of 60° $ (B _{H}=4\times {{10}^{-5}}T) $ is
[MP PET 2005]
Options:
A) 3 A
B) 1.1 A
C) 2.1 A
D) 1.5 A
Show Answer
Answer:
Correct Answer: B
Solution:
$ B=B _{H}\tan \theta $
Therefore $ \frac{{\mu _{0}}ni}{2r}=B _{H}\tan \theta $
Therefore $ i=\frac{2r.\ B _{H}\tan \theta }{{\mu _{0}}n} $
$ =\frac{2\times 0.1\times 4\times {{10}^{-5}}}{10\times 4\pi \times {{10}^{-7}}} $ = 1.1A
BETA
