Test Index

CBSE Class 12 Physics 2015 Delhi Set 1 Paper

© examsnet.com
Question : 24
Total: 26
SECTION - E
(a) State Ampere's circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius ' r ', having ' n ' turns per unit length and carrying a steady current I.
(b) An observer to the left of a solenoid of N turns each of cross section area ' A ' observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m=NIA.
OR
(a) Define mutual inductance and write its S.I. units.
(b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.
(c) In an experiment, two coils c1 and c2 are placed close to each other. Find out the expression for the emf induced in the coil c1 due to a change in the current through the coil c2.
Solution:  
(a) Line integral of magnetic field over a closed loop is equal to the µ0 times the total current passing through the surface enclosed by the loop.

Alternatively
∮
→
B
⋅
→
dl
=µ0I

Let the current flowing through each turn of the toroid be I. The total number of turns equals n . (2πr) where n is the number of turns per unit length. Applying Ampere's circuital law, for the Amperian loop, for interior points.
∮
→
B
⋅
→
dl
=µ0(n2πrl)
B‌dl‌cos‌0=µ0n2πrl
⇒‌‌B×2πr‌=µ0n2πrl
B‌=µ0nI
(b)

The solenoid contains N loops, each carrying a current I. Therefore, each loop acts as a magnetic dipole. The magnetic moment for a current I, flowing in loop of area (vector) A is given by m=IA The magnetic moments of all loops are alignedalong the same direction.
Hence, net magnetic moment equals NIA
OR
(a) ϕ=MI
Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed in the other coil. Alternatively,
e=−M‌
dI
dt

Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.
SI unit: henry / (Weber ampere ‌−1 ) / (volt second ampere ‌−1 )
(b)

Let a current I2 flow through S2 . This sets up a magnetic flux φ1 through each turn of the coil S1 . Total flux linked with S1
N1ϕ1=M12I2 ......(i)
where M12 is the mutual inductance between the two solenoids
Magnetic field due to the current I2 in S2 is µ0n2l2 . Therefore, resulting flux linked with S1.
N1ϕ1=[(n1ℓ)πr12](µ0n2I2) ......(ii)
Comparing (i) & (ii), we get
M12I2‌=(n1ℓ)πr12(µ0n2I2)
∴ M12‌=µ0n1n2πr12ℓ
(c) Let a magnetic flux be (ϕ1) linked with coil C1 due to current (I2) in coil C2 :
We have: ϕ1‌∝I2
⇒ ϕ1‌=MI2
∴ ‌
dϕ1
dt
‌=M‌
dI2
dt

⇒ e‌=−M‌
dI2
dt
© examsnet.com
Go to Question:
Prev Question
Next Question