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  • A coil of 50 turns and area   is pivoted about a vertical

A coil of 50 turns and area  \mathrm{1.25 \times 10^{-2} m^2 } is pivoted about a vertical diameter in a uniform horizontal magnetic field and carries a current of 2A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.04 Nm and when its plane is east-west the corresponding couple is 0.03 Nm. The magnetic induction is K/10 tesla. Find K

Option: 1

2


Option: 2

0


Option: 3

4


Option: 4

6


Answers (1)

best_answer

\mu=\mathrm{NIA}=2 \times 50 \times 1.25 \times 10^{-3}=\frac{1}{8} \mathrm{Am}^2
Given,

\mu \mathrm{B} \cos \theta=0.04 \\

\mu \mathrm{B} \sin \theta=0.03 \\

\mu \mathrm{B}=0.05 \quad \text { (or) } \frac{1}{8} \mathrm{~B}=0.05 \\

\mathrm{~B}=0.4 \text { tesla }=\frac{4}{10} \text { tesla }
 

Posted by

Sayak

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