A coil has an inductance of $50 m H$ and a resistance of $0.3 Ω$ . If $12 V$ emf is applied across the coil, the energy stored in the magnetic field after the current has built up to its steady state value is

40 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

40 m J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 m J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $40 J$
$i_{0} = \frac{E}{R} = \frac{12}{0.3} = 40 A$
$U = \frac{1}{2} L {i_{0}}^{2}$
$= \frac{1}{2} \times 50 \times 10^{- 3} {(40)}^{2} = 40 J$

Suggest Corrections

Similar questions

A coil has an inductance of 5H and resistance

20 Ω

. An emf of

100

v is applied to it. What is the energy stored in the magnetic field, when the current has reached its final value .

A coil having a resistance of 20 ohm and inductance 20 H is connected to 120 V battery. Find the value of energy stored in the magnetic field when the steady state is reached?

Q. A coil of resistance

20 Ω

and inductance

5 H

has been connected to a

100 V

battery The energy stored in the coil after a long time is

Q. A coil having resistance

20 Ω

and inductance

2 H

is connected to a battery of emf

4.0 V

. Find (a) the current at

0.20 s

after the connection is made and (b) the magnetic field energy in the coil at the instant

Q. A coil of resistance

20 Ω

and inductance

5 H

has been connected to a

100 V

battery The energy stored in the coil after a long time is

Join BYJU'S Learning Program

A coil has an inductance of 50mH and a resistance of 0.3Ω. If 12V emf is applied across the coil, the energy stored in the magnetic field after the current has built up to its steady state value is

The correct option is A 40Ji0=ER=120.3=40AU=12Li02=12×50×10−3(40)2=40J

A coil has an inductance of $50 m H$ and a resistance of $0.3 Ω$ . If $12 V$ emf is applied across the coil, the energy stored in the magnetic field after the current has built up to its steady state value is

The correct option is A $40 J$
$i_{0} = \frac{E}{R} = \frac{12}{0.3} = 40 A$
$U = \frac{1}{2} L {i_{0}}^{2}$
$= \frac{1}{2} \times 50 \times 10^{- 3} {(40)}^{2} = 40 J$