An e.m.f. of 5 millivolt is induced in a coil when in a nearby placed another coil, the current changes by 5 ampere in 0.1 second. The coefficient of mutual induction between the two coils will be :

1 Henry

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

0.1 Henry

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

0.1 millihenry

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

0.001 millihenry

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

Open in App

Solution

The correct option is C 0.1 millihenry
Induced emf $E = 5 m V = 0.005 V$
Magnitude of rate of change of current $\frac{d I}{d t} = \frac{5}{0.1} = 50 A s^{- 1}$
Coefficient of mutual induction $M = \frac{E}{\frac{d I}{d t}} = \frac{0.005}{50} = 10^{- 4} H = 0.1 m H$

Suggest Corrections

Similar questions

In a transformer, the coefficient of mutual inductance between the primary and the secondary coil is 0.2 henry. When the current changes by 5 ampere/second in the primary coil, the induced e.m.f. in the secondary coil will be

Q. When the current in a coil changes from 8 ampere to 2 ampere in

3 \times 10^{- 2}

second, the e.m.f. induced in the coil is 2 volt. The self inductance of the coil (in millinery) is

Q. Two coils have a mutual inductance of 5 mH. If the current in the first coil changes by

3 A

0.02

second, then the induced emf in the second coil is

Q. The current decays from

5 A

2 A

0.01 s

in a coil. The emf induced in a coil nearby it is

30 V

. The mutual inductance between the coils is

Q. When the current in a primary coil is changed from

0.5

a m p e r e

2.0

a m p e r e

300

m i l l i s e c o n d s

, then the emf induced in secondary coil is

0.5

v o l t

. Calculate the mutual inductance between the two coils.

Join BYJU'S Learning Program

An e.m.f. of 5 millivolt is induced in a coil when in a nearby placed another coil, the current changes by 5 ampere in 0.1 second. The coefficient of mutual induction between the two coils will be :

The correct option is C 0.1 millihenryInduced emf E=5mV=0.005 VMagnitude of rate of change of current dIdt=50.1=50 A s−1Coefficient of mutual induction M=EdIdt=0.00550=10−4 H=0.1 mH

The correct option is C 0.1 millihenry
Induced emf $E = 5 m V = 0.005 V$
Magnitude of rate of change of current $\frac{d I}{d t} = \frac{5}{0.1} = 50 A s^{- 1}$
Coefficient of mutual induction $M = \frac{E}{\frac{d I}{d t}} = \frac{0.005}{50} = 10^{- 4} H = 0.1 m H$