(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current though it grows from zero to ‘I’.
(b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s−1 till it goes out of the field.
(i) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(ii) Plot a graph showing the variation of magnetic flux and induced emf as a function of time.
OR
(a) Draw the magnetic field lines due to a circular loop area →A carrying current I. Show that it acts as a bar magnet of magnetic moment →m=I→A
(b) Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius‘a’having ’n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distance ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’?
(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current though it grows from zero to ‘I’.
(b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s−1 till it goes out of the field.
(i) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(ii) Plot a graph showing the variation of magnetic flux and induced emf as a function of time.
OR
(a) Draw the magnetic field lines due to a circular loop area →A carrying current I. Show that it acts as a bar magnet of magnetic moment →m=I→A
(b) Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius‘a’having ’n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distance ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’?
(a) Self-inductance of a coil is defined as the ratio of the total flux linked with the coil to the current flowing through it
L=NΦBI
When the current is varied, the flux linked with the coil changes and an e.m.f. is induced in the coil. It is given as
ϵ=−d(NΦB)dt=−LdIdt
The self-induced e.m.f. is also called back e.m.f. as it opposes any change in current in the circuit. So, work needs to be done against back e.m.f. in establishing current. This work done is stored as magnetic potential energy. The rate of doing work is given as
dWdt=|ϵ|=LdIdt (neglecting negative sign) Thus, the total work done in establishing current from 0 to I is
w=∫dW=∫I0LIdI=12LI2
(b)
(i) The direction of induced current in the loop as it goes out is depicted in the figure below
The current will persist till the entire loop comes out of the field.
Hence, we have
t=dv=20cm20cm/s=1s
Hence, the current will persist for 1 second.
(ii) The magnetic flux in the coil when it is inside the field is constant. This maximum flux is given as ϕ
= Bla (a is the side of the square loop). This flux will start dropping once the loop comes out of the field and will be zero when it is completely out of the field.The e.m.f. induced in the coil when it is inside the field is zero as the flux is not changing. When the loop just comes out of the field, the flux change is maximum and the e.m.f. induced is e=−dΦdt=−BIdbdt=−BIv This e.m.f remains constant till the entire loop comes out. When the loop is completely out of the field, the e.m.f. drops to zero again
OR
(a) When a circular loop of area A carries a current I, the loop creates a magnetic field around it. The strength of the magnetic field created depends on the current through the conductor. At the centre of the loop, the magnetic field lines are perpendicular to the plane of the loop which provides the magnetic moment, m = IA. Hence, it behaves like a magnet
The comparison between the magnetic field lines around a current-carrying loop and a bar magnet shows that the two patterns are similar.
If the current in the loop is in the anticlockwise direction, a North Pole is formed and if the current is in the clockwise direction, a South Pole is formed.
(b) Consider a solenoid of length 2l, radius a and having n turns per unit length. It is carrying a current I.
We have to evaluate the axial field at a point P at a distance r from the centre of the solenoid.
Consider a circular element of thickness dx of the solenoid at a distance x from its centre.The magnitude of the field due to this circular loop carrying a current I is given as
dB=μ0dxnla22[(r−x)2+a2]3/2
The magnitude of the total field is obtained by integrating over all the elements from x=−1tox=+1
B=μ0nla22∫+1−1dx2[(r−x)2+a2]3/2
Consider the far axial field of the solenoid, so r ≫a and r≫l. Hence, we have
[(r−x)2+a2]3/2≈r3
So, we get
B=μ0nla22r3∫+1−1dxB=μ0nla22r32I
The magnitude of the magnetic moment of the solenoid is (m×Totalnumberofturns×current×cross−sectionalarea)
m=(n×2I)×I×(πa2)
Therefore, we get the magnetic field as
(μ04π)(2mr3)
This is the expression for magnetic field due to a solenoid on the axial line at a distance r from the centre.This magnetic field is also the field due to a bar magnet of magnetic moment m.
(a) Self-inductance of a coil is defined as the ratio of the total flux linked with the coil to the current flowing through it
L=NΦBI
When the current is varied, the flux linked with the coil changes and an e.m.f. is induced in the coil. It is given as
ϵ=−d(NΦB)dt=−LdIdt
The self-induced e.m.f. is also called back e.m.f. as it opposes any change in current in the circuit. So, work needs to be done against back e.m.f. in establishing current. This work done is stored as magnetic potential energy. The rate of doing work is given as
dWdt=|ϵ|=LdIdt (neglecting negative sign) Thus, the total work done in establishing current from 0 to I is
w=∫dW=∫I0LIdI=12LI2
(b)
(i) The direction of induced current in the loop as it goes out is depicted in the figure below
The current will persist till the entire loop comes out of the field.
Hence, we have
t=dv=20cm20cm/s=1s
Hence, the current will persist for 1 second.
(ii) The magnetic flux in the coil when it is inside the field is constant. This maximum flux is given as ϕ
= Bla (a is the side of the square loop). This flux will start dropping once the loop comes out of the field and will be zero when it is completely out of the field.The e.m.f. induced in the coil when it is inside the field is zero as the flux is not changing. When the loop just comes out of the field, the flux change is maximum and the e.m.f. induced is e=−dΦdt=−BIdbdt=−BIv This e.m.f remains constant till the entire loop comes out. When the loop is completely out of the field, the e.m.f. drops to zero again
OR
(a) When a circular loop of area A carries a current I, the loop creates a magnetic field around it. The strength of the magnetic field created depends on the current through the conductor. At the centre of the loop, the magnetic field lines are perpendicular to the plane of the loop which provides the magnetic moment, m = IA. Hence, it behaves like a magnet
The comparison between the magnetic field lines around a current-carrying loop and a bar magnet shows that the two patterns are similar.
If the current in the loop is in the anticlockwise direction, a North Pole is formed and if the current is in the clockwise direction, a South Pole is formed.
(b) Consider a solenoid of length 2l, radius a and having n turns per unit length. It is carrying a current I.
We have to evaluate the axial field at a point P at a distance r from the centre of the solenoid.
Consider a circular element of thickness dx of the solenoid at a distance x from its centre.The magnitude of the field due to this circular loop carrying a current I is given as
dB=μ0dxnla22[(r−x)2+a2]3/2
The magnitude of the total field is obtained by integrating over all the elements from x=−1tox=+1
B=μ0nla22∫+1−1dx2[(r−x)2+a2]3/2
Consider the far axial field of the solenoid, so r ≫a and r≫l. Hence, we have
[(r−x)2+a2]3/2≈r3
So, we get
B=μ0nla22r3∫+1−1dxB=μ0nla22r32I
The magnitude of the magnetic moment of the solenoid is (m×Totalnumberofturns×current×cross−sectionalarea)
m=(n×2I)×I×(πa2)
Therefore, we get the magnetic field as
(μ04π)(2mr3)
This is the expression for magnetic field due to a solenoid on the axial line at a distance r from the centre.This magnetic field is also the field due to a bar magnet of magnetic moment m.
