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प्रश्न
A current-carrying circular coil of 100 turns and radius 5.0 cm produces a magnetic field of 6.0 × 10−5 T at its centre. Find the value of the current.
उत्तर
Given:
No. of turns, n = 100
Radius of the loop, r = 5 cm = 0.05 m
Magnetic field intensity, B = 6.0 × 10−5 T
Now, let the magnitude of current be i.
\[\text{ Using } B = \frac{\mu_0 \text{ ni} }{2r}, \text{ we get } \]
\[6 . 0 \times {10}^{- 7} = \frac{4\pi \times {10}^{- 7} \times {10}^2 \times i}{2 \times 0 . 05}\]
\[ \Rightarrow i = \frac{60 \times {10}^{- 7}}{4\pi \times {10}^{- 7} \times {10}^2}\]
\[ = 4 . 777 \times {10}^{- 2} \approx 48\] mA
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