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प्रश्न
A current of 1.0 A is established in a tightly wound solenoid of radius 2 cm having 1000 turns/metre. Find the magnetic energy stored in each metre of the solenoid.
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उत्तर
Given:-
Current through the solenoid, i = 1.0 A
Radius of the coil, r = 2 cm
Number of turns per metre, n = 1000
The magnetic energy density is given by \[\frac{B^2}{2 \mu_0}.\]
Volume of the solenoid, V = πr2l
For l = 1m, V = πr2
Thus, the magnetic energy stored in volume V is given by
\[U=\frac{B^2 \pi r^2}{2 \mu_0}\]
The magnetic field is given by
B = μ0ni
= (4π × 10-7) × (1000) × 1
= 4π × 10-4 Τ
\[U = \frac{(4\pi \times {10}^{- 4} )^2 \times 4\pi \times {10}^{- 4}}{2 \times (4\pi \times {10}^{- 7} )} \]
\[ = 8 \pi^2 \times {10}^{- 5} \]
\[ = 78 . 956 \times {10}^{- 5} \]
\[ = 7 . 9 \times {10}^{- 4} J\]
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