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a. Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. b. How does this magnetic energy compare with the electrostatic energy stored in a capacitor? |
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Answer» SOLUTION :a. Work done = Energy spent in ESTABLISHING current 1 in a solenoid is `U_(B)=(1)/(2)LI^(2)` `=(1)/(2)L((B)/(mu_(0)n))` (since `B=mu_(0)nI`, for a solenoid) `=(1)/(2)(mu_(0)n^(2)Al)((B)/(mu_(0)n))^(2)` (from EQ. `6.9.2(3))=(1)/(2mu_(0))B^(2)Al` b. The MAGNETIC energy per UNIT volume V is, `u_(B)=(U_(B))/(V)=(U_(B))/(Al)=(B^(2))/(2mu_(0))` We have already obtained the relation for the square of the field strength. |
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