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Consider a Small Cube of Volume 1 Mm3 at the Centre of a Circular Loop of Radius 10 Cm Carrying a Current of 4 A. Find the Magnetic Energy Stored Inside the Cube.

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प्रश्न

Consider a small cube of volume 1 mm3 at the centre of a circular loop of radius 10 cm carrying a current of 4 A. Find the magnetic energy stored inside the cube.

बेरीज
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उत्तर

Given:-

Current in the loop, i = 4 A

Radius of the loop, r = 10 cm = 0.1 m

Volume of the cube, V = 1 mm3 = 1 × 10-9m

Magnetic field intensity at the centre of the circular loop:-

\[B = \frac{\mu_0 i}{2r}\]

\[ = \frac{(4\pi \times {10}^{- 7} ) \times 4}{2 \times 0 . 1}\]

\[ = 8\pi \times {10}^{- 6} T\]

Magnetic energy density = \[\frac{B^2}{2 \mu_0}\]

otal energy stored in volume V:-

\[U=\frac{B^2 V}{2 \mu_0}\]

\[= \frac{(8\pi \times {10}^{- 6} )^2 \times (1 \times {10}^{- 9} )}{(4\pi \times {10}^{- 7} ) \times 2}\]

\[ = 8\pi \times {10}^{- 14} J\]

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पाठ 38: Electromagnetic Induction - Exercises [पृष्ठ ३१३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 38 Electromagnetic Induction
Exercises | Q 92 | पृष्ठ ३१३

संबंधित प्रश्‍न

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The currents flowing in the two coils of self-inductance L1 = 16 mH and L2 = 12 mH are increasing at the same rate. If the power supplied to the two coil is equal, find the ratio of the energies stored in the two coils at a given instant ?


Derive the expression for the self-inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.


Define self inductance. Write its S.I. units.


Derive an expression for self inductance of a long solenoid of length l, cross-sectional area A having N number of turns.


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(a) self-inductance
(b) rate of Joule heating if the same current goes through them
(c) magnetic field energy if the same current goes through them
(d) time constant if one solenoid is connected to one battery and the other is connected to another battery.


Answer the following question.
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Two coils of wire A and B are placed mutually perpendicular as shown. When current is changed in any one coil.


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Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency `(200/pi)` Hz and a DC source are applied across the coil.

AC Source
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1 3.0 0.5
2 6.0 1.0
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DC Source
S.No. V (volts) I (A)
1 4.0 1.0
2 6.0 1.5
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