If a current increases from zero to one ampere in 0.1 second in a coil of 5 mH, then the magnitude of the induced e.m.f. will be

0.005 volt

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

0.5 volt

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

0.05 volt

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

5 volt

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

Open in App

Solution

The correct option is B 0.05 volt
$ε = L \times (r a t e o f c h a n g e o f c u r r e n t)$
$= (5 \times 10^{- 3}) (1 / 0.1)$
$= 0.05 V$ .

Suggest Corrections

Similar questions

Q. The back e.m.f. induced in a coil, when current changes from 1 ampere to zero in one millisecond, is 4 volts, the self inductance of the coil is

The current in a coil of 0.4 mH coil increases by 250 mA in 0.1 seconds. The induced emf will be:-

Q. The mutual inductance of a pair of coils, each of N turns, is M henry. If a current of I ampere in one of the coils is brought to zero in t second, the emf induced per turn in the other coil, in volt, will be

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.

If a current of 3 amperes flowing in the primary coil is reduced to zero in 0.001second, then the induced e.m.f in the secondary coil is 15000 volts. The mutual inductance between the two coils is

Join BYJU'S Learning Program

If a current increases from zero to one ampere in 0.1 second in a coil of 5 mH, then the magnitude of the induced e.m.f. will be

The correct option is B 0.05 voltε=L×(rate of change of current)=(5×10−3)(1/0.1)=0.05V.

The correct option is B 0.05 volt
$ε = L \times (r a t e o f c h a n g e o f c u r r e n t)$
$= (5 \times 10^{- 3}) (1 / 0.1)$
$= 0.05 V$ .