0 of 88 Questions completed
Questions:
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading…
You must sign in or sign up to start the quiz.
You must first complete the following:
0 of 88 Questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 point(s), (0)
Earned Point(s): 0 of 0, (0)
0 Essay(s) Pending (Possible Point(s): 0)
A square of side \(L\) meters lies in the \(x\) – \(y\) plane in a region, where the magnetic field is given by \(\mathbf{B}=B_o(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \mathrm{T}\), where \(B_o\) is constant. The magnitude of flux passing through the square is
(C) Key concept: Magnetic flux is defined as the total number of magnetic lines of force passing normally through an area placed in a magnetic field and is equal to the magnetic flux linked with that area.
For elementary area \(d A\) of a surface flux linked \(\mathrm{d} \phi=B d A \cos \theta\) or \(d \phi=\vec{B} \cdot d \vec{A}\)
So, Net flux through the surface \(\phi=\oint \vec{B} \times d \vec{A}=B A \cos \theta\)
In this problem, \(A=L^2 \hat{k}\) and \(B=B_0(2 \hat{i}+3 \hat{j}+4 \hat{k}) T\) \(\phi=\vec{B} \cdot \vec{A}=B_0(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot L^2 \hat{k}=4 B_0 L^2 \mathrm{~Wb}\)
A loop, made of straight edges has six corners at \(\mathrm{A}(0,0,0), \mathrm{B}(\mathrm{L}, \mathrm{O}, 0)\) \(\mathrm{C}(L, L, 0), \mathrm{D}(0, L, 0) \mathrm{E}(0, L, L)\) and \(\mathrm{F}(0,0, L)\). A magnetic field \(\mathbf{B}=B_o(\hat{\mathbf{i}}+\hat{\mathbf{k}}) \mathrm{T}\) is present in the region. The flux passing through the loop ABCDEFA (in that order) is
(b) In this problem first we have to analyse area vector, \(\operatorname{loop} A B C D A\) lies in \(x-y\) plane whose area vector \(\vec{A}_1=L^2 \hat{k}\) whereas loop \(A D E F A\) lies in \(y-z\) plane whose area vector \(\overrightarrow{A_2}=L^2 \hat{i}\)
And the magnetic flux is
\(
\begin{aligned}
& \phi_m=\vec{B} \cdot \vec{A} \\
& \vec{A}=\vec{A}_1+\vec{A}_2=\left(L^2 \hat{k}+L^2 \hat{i}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \text { and } \vec{B}=B_0(\hat{i}+\hat{k}) \\
& \text { Now, } \phi_m=\vec{B} \cdot \vec{A}=B_0(\hat{i}+\hat{k}) \cdot\left(L^2 \hat{k}+L^2 \hat{i}\right) \\
& =2 B_0 L^2 \mathrm{~Wb}
\end{aligned}
\)
A cylindrical bar magnet is rotated about its axis (Figure below). A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then
(b) Key concept: The phenomenon of electromagnetic induction is used in this problem. Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes (or a moving conductor cuts the magnetic flux) an emf is produced in the circuit (or emf induces across the ends of the conductor) is called induced emf. The induced emf persists only as long as there is a change or cutting of flux. When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A . Hence the ammeter shows no deflection.
There are two coils \(A\) and \(B\) as shown in Figure below. A current starts flowing in \(B\) as shown, when \(A\) is moved towards \(B\) and stops when \(A\) stops moving. The current in \(A\) is counterclockwise. \(B\) is kept stationary when \(A\) moves. We can infer that
Â
(d) Key concept: Due to variation in the flux linked with coil B an emf will be induced in coil B. Current in coil B becomes zero when coil A stops moving, it is possible only if the current in coil A is constant. If the current in coil A would be variable, there must be some changing flux and then there must be an induced emf. Hence an induced current will be in coil B even when coil A is not moving.
Coil A is made to rotate about a vertical axis (Figure below). No current flows in B if A is at rest. The current in coil A, when the current in B (at \(t=0\) ) is counterclockwise and the coil A is as shown at this instant, \({t}=0\), is
(a) Key concept: In this problem, the Lenz’s law is applicable so let us introduce Lenz’s law first. .
Lenz’s law gives the direction of induced emf/induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon law of conservation of energy.
When the current in coil B (at \({t}=0\) ) is counter-clockwise and the coil A is considered above it. The counter clockwise flow of the current in coil B is equivalent to north pole of magnet and magnetic field lines are eliminating upward to coil A . When coil A starts rotating at \({t}=0\), the current in A is constant along clockwise direction by Lenz’s rule.
The self inductance \(L\) of a solenoid of length \(l\) and area of crosssection \(A\), with a fixed number of turns \(N\) increases as
(b) The self-inductance of a long solenoid of cross-sectional area \(A\) and length \(l\), having \(n\) turns per unit length, filled the inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) is given by
\(
\begin{aligned}
& L=\mu_r \mu_0 n^2 A l \\
& n=N / l
\end{aligned}
\)
where, \({n}={N} / {l}\) (no. of turns per unit length)
1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross section: Larger the area of cross section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases \(\mu_{r}\) times if it is wound over an iron core of relative permeability \(\mu_{r}\).
The long solenoid of cross-sectional area A and length \(l\), having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is \(L=\mu_{r} \mu_0 \mathrm{~N}^2 \mathrm{~A} / \mathrm{l}\)
So, the self inductance \(L\) of a solenoid increases as \(l\) decreases and \(A\) increases because \(L\) is directly proportional to area and inversely proportional to length.
Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.
A metal plate is getting heated. It can be because
(a, b, d) A metal plate is getting heated when a DC or AC current is passed through the plate, known as heating effect of current. This current (called eddy current) is induced in the plate when a metal plate is subjected to a time varying magnetic field, i.e., the magnetic flux linked with the plate changes and eddy currents comes into existence which make the plate hot.
An e.m.f is produced in a coil, which is not connected to an external voltage source. This can be due to
(a, b, c) Key concept: As we know whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes, an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.In this problem, magnetic flux linked with the isolated coil changes when the coil is placed in the region of a time varying magnetic field, the coil moving in a constant magnetic field or in time varying magnetic field.
The mutual inductance \(M_{12}\) of coil 1 with respect to coil 2
(a, d) Key concept: Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.
The mutual inductance \(M_{12}\) of coil 1 w.r.t. coil 2 increases when they are brought nearer and is the same as \(M_{21}\) of coil 2 with respect to coil 1.
\(M_{12}\), i.e., mutual inductance of solenoid \(S_1\) with respect to solenoid \(S_2\) is given by
\(
M_{12}=\frac{\mu_0 N_1 N_2 \pi r_1^2}{l}
\)
where signs are as ùsual.
Also, \(M_{21}\), i.e., mutual inductance of solenoid \(S_2\) with respect to solenoid \(S_1\) is given by
\(
M_{21}=\frac{\mu_0 N_1 N_2 \pi r_1^2}{l}
\)
So, we have \(M_{12}=M_{21}=M\)
A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because
(b, c) Key concept: As we know whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
The induced emf is given by rate of change of magnetic flux linked with the circuit, i.e., \(e=-d \Phi / d t\) According to the problem there is no electromotive force produced in the coil. Then the various arrangement are to be thought of in such a way that the magnetic flux linked with the coil does not change even if the coil is placed and expanded in magnetic field.
When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or we can say that direction of magnetic field is perpendicular to the direction of area (increasing) so that their dot product is always zero and hence change in magnetic flux is also zero.
Or, The magnetic field has a perpendicular (to the plane of the coil) component whose-magnitude is decreasing suitably in such a way that the dot product of magnetic field and surface area of plane of coil remain constant at every instant.
A rod of length \(l\) rotates with a small but uniform angular velocity \(\omega\) about its perpendicular bisector. A uniform magnetic field \(B\) exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is
(b) The motional emf induced in a small element of length \(d x\) at a distance \(x\) from the center is \(d(\Delta \boldsymbol{V})=\boldsymbol{B} \omega x d x\).
The total potential difference is the integral of the induced emf over the length of the rod.
The induced EMF in a small element of length \(d x\) at a distance \(x\) from the center is given by:
\(d(\Delta V)=B \cdot v \cdot d x\)
Where \(v\) is the linear velocity of the element, which is \(v=\omega x\).
Therefore,
\(d(\Delta V)=B \omega x d x\)
The total potential difference \(\Delta V\) between the center and one end is found by integrating \(d(\Delta V)\) from 0 to \(\frac{l}{2}\) :
\(\Delta V=\int_0^{\frac{l}{2}} B \omega x d x\)
\(\Delta V=B \omega \int_0^{\frac{l}{2}} x d x\)
\(\Delta V=B \omega\left[\frac{x^2}{2}\right]_0^{\frac{l}{2}}\)
\(\Delta V=B \omega\left(\frac{\left(\frac{1}{2}\right)^2}{2}-\frac{0^2}{2}\right)\)
\(\Delta V=B \omega \frac{l^2}{8}\)
\(\Delta V=\frac{1}{8} B \omega l^2\)
The potential difference between the center of the rod and an end is \(\frac{1}{8} \omega B l^2\).
A rod of length \(l\) rotates with a uniform angular velocity \(\omega\) about its perpendicular bisector. A uniform magnetic field \(B\) exists parallel to the axis of rotation. The potential difference between the two ends of the rod is
We have a rod of length \(L\) rotating about its perpendicular bisector with an angular velocity \(\omega\).
A uniform magnetic field \(B\) exists parallel to the axis of rotation.
The linear velocity \(v\) of a point at a distance \(x\) from the center of the rod is given by:
\(
v=\omega x
\)
Take a small strip of the rod at a distance \(x\) from the center with a width \(d x\).
The induced electromotive force (emf) \(d E\) across this strip can be expressed as:
\(
d E=B \cdot v \cdot d x=B \cdot(\omega x) \cdot d x
\)
To find the total potential difference \(E\) between the two ends of the rod, integrate \(d E\) from \(-\frac{L}{2}\) to \(\frac{L}{2}\) :
\(E=\int_{-\frac{L}{2}}^{\frac{L}{2}} B \omega x d x\)
Factor out constants \(B\) and \(\omega\) :
\(
E=B \omega \int_{-\frac{L}{2}}^{\frac{L}{2}} x d x
\)
The integral of \(x\) is:
\(
\int x d x=\frac{x^2}{2}
\)
Evaluating the integral from \(-\frac{L}{2}\) to \(\frac{L}{2}\) :
\(
E=B \omega\left(\frac{x^2}{2}\right)_{-\frac{L}{2}}^{\frac{L}{2}}=B \omega\left(\frac{\left(\frac{L}{2}\right)^2}{2}-\frac{\left(-\frac{L}{2}\right)^2}{2}\right)
\)
Both terms yield the same value, thus:
\(
E=B \omega\left(\frac{L^2}{8}-\frac{L^2}{8}\right)=0
\)
The potential difference \(E\) between the two ends of the rod is:
\(
E=0
\)
Consider the situation shown in the figure below. If the switch is closed and after some time it is opened again, the closed loop will show
(d) When the switch is closed, the current will flow in the downward direction in part AB of the circuit nearest to the closed loop.
Due to current in wire AB , a magnetic field will be produced in the loop. This magnetic field due to increasing current will be the cause of the induced current in the closed loop. According to Lenz’s law, the induced current is such that it opposes the increase in the magnetic field that induces it. So, the induced current will be in clockwise direction opposing the increase in the magnetic field in upward direction.
Similarly, when the circuit is opened, the current will suddenly fall in the circuit, leading to decrease in the magnetic field in the loop. Again, according to Lenz’s law, the induced current is such that it opposes the decrease in the magnetic field. So, the induced current will be in anti-clockwise direction, opposing the decrease in the magnetic field in upward direction.
Consider the situation shown in the figure. If the closed loop is completely enclosed in the circuit containing the switch, the closed loop will show
(c)Â According to Lenz’s law, the induced current in the loop will be such that it opposes the increase in the magnetic field due to current flow in the circuit. Therefore, the direction of the induced current when the switch is closed is anti-clockwise.
Similarly, when the switch is open, there is a sudden fall in the current, leading to decrease in the magnetic field at the centre of the loop. According to Lenz’s law, the induced current in the loop is such that it opposes the decrease in the magnetic field. Therefore, the direction of the induced current when the switch is open is clockwise.
A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time the magnet
(b) As the magnet is moving under gravity, the flux linked with the copper tube will change because of the motion of the magnet. This will produce eddy currents in the body of the copper tube. According to Lenz’s law, these induced currents oppose the fall of the magnet. So, the magnet will experience a retarding force. This force will continuously increase with increasing velocity of the magnet till it becomes equal to the force of gravity. After this, the net force on the magnet will become zero. Hence, the magnet will attain a constant speed.
The figure below shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will
(c) For the circuit,
\(
E=-L \frac{d i}{d t}
\)
The current will increase in the solenoid, flowing in clockwise direction in the circuit. Due to this increased current, the flux linked with the copper ring with increase with time, causing an induced current. This induced current will oppose the cause producing it. Hence, the current in the copper ring will be in anticlockwise direction. Now, because the directions of currents in the solenoid and ring are opposite, the ring will be repelled and hence will move away from the solenoid.
Consider the following statements:
(A) An emf can be induced by moving a conductor in a magnetic field.
(B) An emf can be induced by changing the magnetic field.
(a) Statement A is true, as an emf can be induced by moving a conductor with some velocity \(v\) in a magnetic field \(B\). It is given by
\(
e=B v l
\)
Statement B is true, as an emf can be induced by changing the magnetic field that causes the change in flux \(\phi\) through a conductor or a loop. It is given by
\(
e=-\frac{d \phi}{d t}
\)
Consider the situation shown in the figure below. The wire \(A B\) is slid on the fixed rails with a constant velocity. If the wire \(A B\) is replaced by a semicircular wire, the magnitude of the induced current will
(b)
The induced emf across ends \(A\) and \(B\) is given by
\(
E=B v l
\)
This induced emf will serve as a voltage source for the current to flow across resistor \(R\), as shown in the figure. The direction of the current is given by Lenz’s law and it is anticlockwise.
\(
i=\frac{B v l}{R}
\)
If the wire is replaced by a semicircular wire, the induced current will remain the same, as it depends on the length of the wire and not on its shape (when \(B, v\) and \(R\) are kept constant).
Figure-a below shows a conducting loop being pulled out of a magnetic field with a speed \(v\). Which of the four plots shown in figure-b below may represent the power delivered by the pulling agent as a function of the speed \(v\) ?
(b) The emf developed across the ends of the loop is given by
\(
e=B v l
\)
If \(R\) is the resistance of the loop, then the power delivered to the loop is given by
\(
\begin{aligned}
& P=\frac{e^2}{R}=\frac{B^2 v^2 l^2}{R} \\
& \Rightarrow P \propto v^2
\end{aligned}
\)
This relation is best represented by plot \(b\) in the figure.
Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. During this period, the two loops
(a) The two loops will attract each other. Since the current in the first loop is increasing due to the decreasing resistance with temperature, the induced current in the second loop will be in the same direction, leading to attraction.
Explanation:
1. Magnetic Field:
When current flows in a loop, it generates a magnetic field. The direction of this field can be determined using the right-hand rule.
2. Induced Current:
When the current in the first loop changes, it creates a changing magnetic flux through the second loop. This induces a current in the second loop.
3. Lenz’s Law:
The direction of the induced current opposes the change in magnetic flux. In this case, the increasing current in the first loop causes the flux through the second loop to increase. To oppose this, the induced current in the second loop will flow in the same direction as the current in the first loop.
4. Attraction:
Since the currents in both loops are in the same direction, they will attract each other.Â
A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is
(c) The magnetic field inside the solenoid is parallel to its axis. If the plane of the loop contains the axis of the solenoid, then the angle between the area vector of the circular loop and the magnetic field is zero. Thus, the flux through the circular loop is given by
\(
\phi=B A \cos \theta=B A \cos 0^{\circ}=B A
\)
Here,
\(B=\) Magnetic field due to the solenoid
\(A=\) Area of the circular loop
\(\theta=\) Angle between the magnetic field and the area vector
Now, the induced emf is given by
\(
\begin{aligned}
& e=-\frac{d \phi}{d t} \\
& \because \phi=B A=\mathrm{constant} \\
& \therefore {e}=0
\end{aligned}
\)
We can see that the induced emf does not depend on the varying current through the solenoid and is zero for constant flux through the loop. Because there is no induced emf, no current is induced in the loop.
A conducting square loop of side \(l\) and resistance \(R\) moves in its plane with a uniform velocity \(v\) perpendicular to one of its sides. A uniform and constant magnetic field \(B\) exists along the perpendicular to the plane of the loop as shown in the figure below. The current induced in the loop is
(d)
Figure (a) shows the square loop moving in its plane with a uniform velocity \(v\).
Figure (b) shows the equivalent circuit.
The induced emf across ends AB and CD is given by
\(
E=B v l
\)
On applying KVL in the equivalent circuit, we get
\(
\begin{aligned}
& E-E+i R=0 \\
& \Rightarrow i=0
\end{aligned}
\)
No current will be induced in the circuit due to zero potential difference between the closed ends.
A bar magnet is moved along the axis of a copper ring placed far away from the magnet. Looking from the side of the magnet, an anticlockwise current is found to be induced in the ring. Which of the following may be true?
(b, c)Â It can be observed that the induced current is in an anti-clockwise direction. So, the magnetic field induced in the copper ring is towards the observer.
According to Lenz’s law, the current induced in a circuit due to a change in the magnetic flux is in such direction so as to oppose the change in flux.
Two cases are possible:-
(1) The magnetic flux is increasing in the direction from the observer to the circular coil.
(2) The magnetic flux is decreasing in the direction from the coil to the observer.
So, from the above-mentioned points, the following conclusions can be made:-
1. The south pole faces the ring, and the magnet moves away from it.
2. The north pole faces the ring, and the magnet moves towards it.
A conducting rod is moved with a constant velocity \(v\) in a magnetic field. A potential difference appears across the two ends
(d) The potential difference across the two ends is given by
\(
e=B v l
\)
It is non-zero only
(i) if the rod is moving in the direction perpendicular to the magnetic field \((\vec{v} \perp \vec{B})\)
(ii) if the velocity of the rod is in the direction perpendicular to the length of the \(\operatorname{rod}(\vec{v} \perp \vec{l})\)
(iii) if the magnetic field is perpendicular to the length of the \(\operatorname{rod} \vec{l} \perp \vec{B}\)
Thus, none of the above conditions is satisfied in the alternatives given.
A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if
(c, d) An emf will be induced in the loop if it is rotated about a diameter or if it is deformed. Rotating the loop about its axis will not induce an emf because the magnetic flux through the loop remains constant. Similarly, translating the loop in a uniform magnetic field does not change the magnetic flux, so no emf is induced. –
A metal sheet is placed in front of a strong magnetic pole. A force is needed to
(a) hold the sheet there if the metal is magnetic
(c) move the sheet away from the pole with uniform velocity if the metal is magnetic
(d) move the sheet away from the pole with uniform velocity if the metal is nonmagnetic
The strong magnetic pole will attract the magnet, so a force is needed to hold the sheet there if the metal is magnetic.
If we move the metal sheet (magnetic or nonmagnetic) away from the pole, eddy currents are induced in the sheet. Because of eddy currents, thermal energy is produced in it. This energy comes at the cost of the kinetic energy of the plate; thus, the plate slows down. So, a force is needed to move the sheet away from the pole with uniform velocity.
A constant current \(i\) is maintained in a solenoid. Which of the following quantities will increase if an iron rod is inserted in the solenoid along its axis?
(a) magnetic field at the centre
(b) magnetic flux linked with the solenoid
(c) self-inductance of the solenoid
Iron rod has high permeability. When it is inserted inside a solenoid the magnetic field inside the solenoid increases. As magnetic field increases inside the solenoid thus the magnetic flux also increases. The Selfinductance (L) of the coil is directly proportional to the permeability of the material inside the solenoid. As the permeability inside the coil increases. Therefore, the self-inductance will also increase.
Two solenoids have identical geometrical construction but one is made of thick wire and the other of thin wire. Which of the following quantities are different for the two solenoids?
(b, d) The quantities that are different for the two solenoids are (b) rate of Joule heating and (d) time constant.
Because the solenoids are identical, their self-inductance will be the same.
Resistance of a wire is given by
\(
R=\rho \frac{l}{A}
\)
Here,
\(l\)= Length of the wire
\(A=\) Area of cross section of the wire
\(\rho=\) Resistivity of the wire
Because \(\rho\) and 1 are the same for both wires, the thick wire will have greater area of cross section and hence less resistance than the thin wire.
\(
\Rightarrow R_{\text {thick }}<R_{\text {thin }}
\)
The time constant for a solenoid is given by
\(
\begin{aligned}
& \tau=\frac{L}{R} \\
& \therefore \tau_{\text {thick }}>\tau_{\text {thin }}
\end{aligned}
\)
Thus, time constants of the solenoids would be different if one solenoid is connected to one battery and the other is connected to another battery.
Also, because the self-inductance of the solenoids is the same and the same current flows through them, the magnetic field energy given by \(\frac{1}{2} L i^2\) will be the same.
Power dissipated as heat is given by
\(
P=i^2 R
\)
\(i\) is the same for both solenoids.
\(
\therefore P_{\text {thick }}<P_{\text {thin }}
\)
Because the resistance of the coils are different, the rate of Joule heating will be different for the coils if the same current goes through them.
An \(L R\) circuit with a battery is connected at \(t=0\). Which of the following quantities is not zero just after the connection?
(d) At time \(\mathrm{t}=0\), the current in the L-R circuit is zero. The magnetic field energy is given by \(U=\frac{1}{2} L i^2\), as the current is zero the magnetic field energy will also be zero. Thus, the power delivered by the battery will also be zero. As, the LR circuit is connected to the battery at \(t=0\), at this time the current is on the verge to start growing in the circuit. So, there will be an induced emf in the inductor at the same time to oppose this growing current.
A \(\operatorname{rod} A B\) moves with a uniform velocity \(v\) in a uniform magnetic field as shown in the figure below.
(b) The end A becomes positively charged
Due to electromagnetic induction, emf \(e\) is induced across the ends of the rod. This induced emf is given by
\(
e=B v l
\)
The direction of this induced emf is from A to B , that is, A is at the higher potential and B is at the lower potential. This is because the magnetic field exerts a force equal to \(q v B\) on each free electron where q is \(-1.6 \times 10^{-16} \mathrm{C}\). The force is towards AB by Fleming’s left-hand rule; hence, negatively charged electrons move towards the end \(B\) and get accumulated near it. So, a negative charge appears at \(B\) and a positive charge appears at A.
\(L, C\) and \(R\) represent the physical quantities inductance, capacitance and resistance respectively. Which of the following combinations have dimensions of frequency?
(a, b, c) The time constant of the RC circuit is given by
\(
\tau=R C
\)
On taking the reciprocal of the above relation, we get
\(
f_1=\frac{1}{R C} \dots(1)
\)
\(\mathrm{f}_1\) will have the dimensions of the frequency.
The time constant of the LR circuit is given by
\(
\tau=\frac{L}{R}
\)
On taking the reciprocal of the above relation, we get
\(
f_2=\frac{R}{L} \dots(2)
\)
\(f_2\) will have the dimensions of the frequency.
On multiplying eq. (1) and (2), we get
\(
f_1 f_2=\frac{1}{L C}
\)
\(
\begin{aligned}
&\Rightarrow \sqrt{f_1 f_2}=\frac{1}{\sqrt{L C}}\\
&\text { Thus, } \sqrt{f_1 f_2} \text { will have the dimensions of the frequency. }
\end{aligned}
\)
The switches in figure (a) and (b) are closed at \(t=0\) and reopened after a long time at \(t=t_0\).
(b) The charge on \(C\) long after \(\mathrm{t}=0\) is \(\varepsilon \mathrm{C}\).
(c) The current in \(L\) just before \(\mathrm{t}=\mathrm{t}_0\) is \(\varepsilon / \mathrm{R}\).
The charge on the capacitor at time ” t ” after connecting it with a battery is given by,
\(
Q=C \varepsilon\left[1-e^{-t / \operatorname{RC}}\right]
\)
Just after \(\mathrm{t}=0\), the charge on the capacitor will be
\(
Q=C \varepsilon\left[1-e^0\right]=0
\)
For a long after time, \(\mathrm{t} \rightarrow \infty\)
Thus, the charge on the capacitor will be
\(
\begin{aligned}
& Q=C \varepsilon\left[1-e^{-\infty}\right] \\
& \Rightarrow Q=C \varepsilon[1-0]=C \varepsilon
\end{aligned}
\)
The current in the inductor at time ” t ” after closing the switch is given by
\(
I=\frac{V_b}{R}\left(1-e^{-t R / L}\right)
\)
Just before the time \(\mathrm{t}_0\), current through the inductor is given by
\(
I=\frac{V_b}{R}\left(1-e^{-t_0 R L}\right)
\)
It is given that the time \(t_0\) is very long.
\(
\begin{aligned}
& \therefore \mathrm{t}_0 \rightarrow \infty \\
& I=\frac{\varepsilon}{R}\left(1-e^{-\infty}\right)=\frac{\varepsilon}{R}
\end{aligned}
\)
When the switch is opened, the current through the inductor after a long time will become zero.
The magnetic flux linked with a vector area \(\mathbf{A}\) in a uniform magnetic field \(\mathbf{B}\) is
(c)
\(
\begin{aligned}
&\text { The formula for magnetic flux }\left(\boldsymbol{\Phi}_B\right) \text { is: }\\
&\Phi_B=\mathbf{B} \cdot \mathbf{A}
\end{aligned}
\)
The unit of magnetic flux is
(b) \(\text { The correct unit of magnetic flux is the weber, } \mathrm{Wb} \text {. }\)
The dimensions of magnetic flux are
(d) Dimensions of \(\boldsymbol{B}:\left[\mathrm{MT}^{-2} \mathrm{~A}^{-1}\right]\)
Dimensions of \(A:\left[\mathrm{L}^2\right]\)
Dimensions of \(\Phi_B=\left[\mathrm{MT}^{-2} \mathrm{~A}^{-1}\right] \times\left[\mathrm{L}^2\right]\)
The dimensional formula of magnetic flux is \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\).
The magnetic flux linked with a coil, in weber is given by the equation \(\phi=3 t^2+4 t+9\). Then, the magnitude of induced emf at \(t=2 \mathrm{~s}\) will be
(d) Magnitude of induced emf,
\(
\begin{aligned}
e & =-\frac{d \phi}{d t} \\
& =-\frac{d}{d t}\left(3 t^2+4 t+9\right)=-(6 t+4)
\end{aligned}
\)
At \(t=2 \mathrm{~s}, \quad e=-[6 \times(2)+4]=-16\)
\(
\therefore \quad|e|=16 \mathrm{~V}
\)
A coil having an area \(A_0\) is placed in a magnetic field which changes from \(B_0\) to \(4 B_0\) in time interval \(t\). The emf induced in the coil will be
\(
\text { (a) As, }|e|=\frac{\Delta \phi}{\Delta t}=\frac{4 B_0 A_0-B_0 A_0}{t}=\frac{3 B_0 A_0}{t}
\)
The magnetic flux \(\phi\) (in weber) in a closed circuit of resistance \(10 \Omega\) varies with time \(t\) (in second) according to equation \(\phi=6 t^2-5 t+1\). The magnitude of induced current at \(t=0.25 \mathrm{~s}\) is
(d) Induced emf, \(e=\left|\frac{d \phi}{d t}\right|=|12 t-5|\)
At \(t=0.25 \mathrm{~s}, \Rightarrow e=2 \mathrm{~V}\)
\(\therefore\) Induced current, \(i=\frac{e}{R}=0.2 \mathrm{~A}\)
The magnetic flux across a loop of resistance \(10 \Omega\) is given by \(\phi=\left(5 t^2-4 t+1\right) \mathrm{Wb}\). How much current is induced in the loop after 0.2 s?
(b) Magnetic flux, \(\phi=\left(5 t^2-4 t+1\right) \mathrm{Wb}\)
\(
\therefore \quad \frac{d \phi}{d t}=(10 t-4) \mathrm{Wbs}^{-1}
\)
The induced emf, \(e=\frac{-d \phi}{d t}=-(10 t-4)\)
At \(t=0.2 \mathrm{~s}, e=-(10 \times 0.2-4)=2 \mathrm{~V}\)
The induced current, \(I=\frac{e}{R}=\frac{2 \mathrm{~V}}{10 \Omega}=0.2 \mathrm{~A}\)
A circular ring of diameter 20 cm has a resistance \(0.01 \Omega\). How much charge will flow through the ring, if it is rotated from a position perpendicular to a uniform magnetic field of \(B=2 \mathrm{~T}\) to a position parallel to field?
\(
\text { (b) As, charge, }|\Delta q|=\frac{\Delta \phi}{R}=\frac{B A}{R}=\frac{(2)\left(\frac{\pi}{4}\right)(0.2)^2}{0.01}=6.28 \mathrm{C}
\)
In a circuit with a coil of resistance \(2 \Omega\), the magnetic flux changes from 2.0 Wb to 10.0 Wb in 0.2 s. The charge that flows in the coil during this time is
\(
\text { ( b) As, charge, } \quad \Delta Q=\frac{\Delta \phi}{R}=\frac{\phi_2-\phi_1}{R}=\frac{(10-2)}{2}=4 \mathrm{C}
\)
The direction of induced emf during electromagnetic induction is given by
(b) Lenz’s law.
Explanation:
Lenz’s law states that the direction of an induced electromotive force (emf) is such that the magnetic field created by the induced current opposes the change in the original magnetic field. In simpler terms, it means the induced current will always try to resist the change in magnetic flux.
Two different loops are concentric and lie in the same plane. The current in the outer loop is clockwise and increasing with time. The induced current in the inner loop, is
(c) According to Lenz’s law, the induced current will be in such a direction, so that it opposes the change due to which it is produced. So, the induced current in the inner loop will be in counter-clockwise direction.
When the current through a solenoid increases at a constant rate, then the induced current
(b) When the current through a solenoid increases at a constant rate, the induced current is a constant and is opposite to the direction of the inducing current.
Explanation:
According to Lenz’s Law, the induced current in a circuit opposes the change in magnetic flux that produced it. When the current in a solenoid increases at a constant rate, the magnetic field within the solenoid also increases, and the magnetic flux through the solenoid changes. This changing magnetic flux induces a current in the solenoid that opposes the increase in the magnetic field, according to physics concepts.
The north pole of a long horizontal bar magnet is being brought closer to a vertical conducting plane along the perpendicular direction. The direction of the induced current in the conducting plane will be
(d) According to Lenz’s law, the direction of induced current is such as to attain north polarity. So, it will be anti-clockwise.
There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. Then,
(a) Due to change in the shape of the loop, the magnetic flux linked with the loop increases. Hence, current is induced in the loop in such a direction that it opposes the increase in flux. Therefore, induced current flows in the anti-clockwise direction.
A conducting rod of length \(l\) is falling with a constant velocity \(v\) perpendicular to a uniform horizontal magnetic field \(B\). A potential difference between its two ends will be
(b) The formula for motional EMF \((\varepsilon)\) is given by:
\(
\varepsilon=B l v
\)
where \(\boldsymbol{B}\) is the magnetic field strength, \(l\) is the length of the conductor, and \(v\) is the velocity of the conductor.
A wire of length 50 cm moves with a velocity of \(300 \mathrm{~m} / \mathrm{min}\), perpendicular to a magnetic field. If the emf induced in the wire is 2 V , then the magnitude of the field in tesla is
(d) When a wire of length \(l\) moves with velocity \(v\), perpendicular to a magnetic field \(B\), then the emf is induced.
The magnitude of induced emf is given by \(|\varepsilon|=B l v\).
Given, \(l=50 \mathrm{~cm}=0.5 \mathrm{~m}, v=300 \mathrm{~m} / \mathrm{min}=5 \mathrm{~m} / \mathrm{s}\)
\(
|\varepsilon|=2 \mathrm{~V} \text { or } B=\frac{|\varepsilon|}{l v}=\frac{2}{0.5 \times 5}=0.8 \mathrm{~T}
\)
A 10 m wire kept in east-west direction is falling with velocity \(5 \mathrm{~m} / \mathrm{s}\) perpendicular to the field \(0.3 \times 10^4 \mathrm{~Wb} / \mathrm{m}^2\). The induced emf across the terminal will be
(b)
\(
\begin{aligned}
\text { Induced emf } & =B l v=0.3 \times 10^{-4} \times 10 \times 5 \\
& =1.5 \times 10^{-3} \mathrm{~V}=1.5 \mathrm{mV}
\end{aligned}
\)
A boat is moving due east in a region, where the earth’s magnetic field is \(5.0 \times 10^{-5} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\) due to north and horizontal. The boat carried a vertical aerial 2 m long. If the speed of the boat is \(1.50 \mathrm{~ms}^{-1}\), then the magnitude of the induced emf in the wire of aerial is
(b) Induced emf
\(
\begin{aligned}
& =B \times v \times l \\
& =5.0 \times 10^{-5} \times 1.50 \times 2 \\
& =10.0 \times 10^{-5} \times 1.5 \\
& =1.5 \times 10^{-4}=0.15 \mathrm{mV}
\end{aligned}
\)
A 0.1 m long conductor carrying a current of 50 A is held perpendicular to a magnetic field of 1.25 mT. The mechanical power required to move the conductor with a speed of \(1 \mathrm{~ms}^{-1}\) is
(c) Here,
\(
\begin{aligned}
& l=0.1 \mathrm{~m}, v=1 \mathrm{~ms}^{-1} \\
& I=50 \mathrm{~A} \text { and } B=1.25 \mathrm{mT}=1.25 \times 10^{-3} \mathrm{~T}
\end{aligned}
\)
The induced emf, \(\varepsilon=B l v\)
The mechanical power,
\(
\begin{aligned}
P & =\varepsilon I=B l v I=1.25 \times 10^{-3} \times 0.1 \times 1 \times 50 \\
& =6.25 \times 10^{-3} \mathrm{~W}=6.25 \mathrm{~mW}
\end{aligned}
\)
A conducting rod of length \(l\) is moving in a transverse magnetic field of strength \(B\) with velocity \(v\). The resistance of the rod is \(R\). The current in the rod is
(c) Since the rod is moving in transverse magnetic field, so it will cut no flux passing through the field and hence no induced emf is produced. So, no current will flow through the rod.
A metallic square loop \(A B C D\) is moving in its own plane with velocity \(v\) in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced
(d) The square loop \(A B C D\) is a metallic loop. So it consists of the free electrons. When the loop moves in a magnetic field, the free electrons in the AD and BC rod will experience the Lorentz force and they will move towards the ends creating a potential difference in the rods. This induced emf leads to the electric field induced in the rods AD and BC of the square loop. But there will be no current flowing in the loop as ned potential difference is zero.
A coil of \(N\) turns and mean cross-sectional area \(A\) is rotating with uniform angular velocity \(\omega\) about an axis at right angle to uniform magnetic field \(B\). The induced emf \(E\) in the coil will be
(d) The induced emf \(E\) in the coil is given by \(E=N B A \omega \sin (\omega t)\), where \(N\) is the number of turns, \(B\) is the magnetic field strength, \(A\) is the cross-sectional area of the coil, \(\omega\) is the angular velocity, and \(t\) is time.
Explanation:
The magnetic flux \(\Phi\) through the coil is given by \(\Phi=N B A \cos (\theta)\), where \(\theta\) is the angle between the magnetic field and the normal to the coil. Since the coil is rotating with angular velocity \(\omega\), the angle \(\theta\) can be expressed as \(\theta=\omega t\). Therefore, the flux is \(\Phi=N B A \cos (\omega t)\).
According to Faraday’s law of induction, the induced emf \(E\) is the negative time derivative of the magnetic flux:
\(
E=-\frac{d \Phi}{d t}=-\frac{d}{d t}(N B A \cos (\omega t))=N B A \omega \sin (\omega t)
\)
Thus, the induced emf is proportional to \(N B A \omega \sin (\omega t)\).
A circular coil of mean radius of 7 cm and having 4000 turns is rotated at the rate of \(1800 \mathrm{rev} / \mathrm{min}\) in the earth’s magnetic field ( \(B=0.5 \mathrm{G}\) ), the maximum emf induced in coil will be
\(
\begin{aligned}
& \text { (a) Induced emf, } e_0=\omega N B A=(2 \pi v) N B\left(\pi r^2\right)=2 \times \pi^2 v N B r^2 \\
& =2 \times(3.14)^2 \times \frac{1800}{60} \times 4000 \times 0.5 \times 10^{-4} \times\left(7 \times 10^{-2}\right)^2=0.57 \mathrm{~V}
\end{aligned}
\)
A metal rod of length 2 m is rotating with an angular velocity of \(100 \mathrm{rads}^{-1}\) in a plane perpendicular to a uniform magnetic field of 0.3 T. The potential difference between the ends of the rod is
(d) Induced emf, when rod rotates in a vertical plane perpendicular to magnetic field is given as
\(
e=\frac{1}{2} B l^2 \omega=\frac{1}{2} \times 0.3 \times 2^2 \times 100=60 \mathrm{~V}
\)
A rectangular coil rotates about an axis normal to the magnetic field. If \(E_m\) is the maximum value of the induced emf, then the instantaheous emf when the plane of the coil makes an angle of \(45^{\circ}\) with the magnetic field is
(c) \(E=E_m \cos 45^{\circ}\)
where, \(E_m\) is the maximum value of induced emf.
\(
\therefore \quad E=\frac{E_m}{\sqrt{2}}
\)
Consider the following statements.
A. An emf can be induced by moving a conductor in a magnetic field.
B. A magnetic field can be produced by changing the electric field.
(a) Both \(A\) and \(B\) are true.
Explanation:
Statement A is true:
This is a fundamental principle of electromagnetic induction. When a conductor moves through a magnetic field, a changing magnetic flux through the conductor induces an electromotive force (EMF) within the conductor.
Statement B is true:
This is also a fundamental principle of electromagnetism. Faraday’s law of induction states that a changing magnetic field can induce an EMF in a closed loop. Conversely, a changing electric field creates a magnetic field.
The SI unit of inductance, can be written as
\(
\text { (b) } U=\frac{1}{2} L i^2,[L]=\left[\frac{U}{i^2}\right]=\frac{\text { joule }}{\text { ampere }^2}
\)
\(
\begin{aligned}
&\begin{aligned}
& {[L]=\left[\frac{e}{d i / d t}\right]=\text { ohm-second }} \\
& {[L]=\left[\phi_B / i\right]=\text { weber/ampere }} \\
& {[L]=[\phi / i]=[e d t / d i]=\frac{\text { volt-second }}{\text { ampere }}}
\end{aligned}\\
&\text { Hence, SI unit of inductance is volt-second/ampere. }
\end{aligned}
\)
A long solenoid has 500 turns. When a current of 2 A is passed through it, then the resulting magnetic flux linked with each turn of the solenoid is \(4 \times 10^{-3} \mathrm{~Wb}\). The self-inductance of the solenoid is
(a) We have, \(N \phi_B=L i\)
Self-inductance, \(L=\frac{N \phi_B}{i}=\frac{500 \times 4 \times 10^{-3}}{2}=1 \mathrm{H}\)
If a current of 10 A changes in one second through a coil, and the induced emf is 10 V ,then the self-inductance of the coil is
(d) Self-inductance of coil, \(|e|=L \frac{d i}{d t}\)
\(
10=L \times \frac{10}{1}
\)
\(\therefore\) Self-inductance, \(L=1 \mathrm{H}\)
During a current change from 2 A to 4 A in \(0.5 \mathrm{~s}, 8 \mathrm{~V}\) of emf is developed in a coil. The coefficient of self-induction is
\(
\text { (b) Self-inductance, } L=\left|\frac{e}{d i / d t}\right|=\frac{8}{(2 / 0.5)}=2 \mathrm{H}
\)
The current passing through a choke coil of 5 H is decreasing at the rate of \(2 \mathrm{~A} / \mathrm{s}\). The emf developing across the coil is
(a) Given, \(\frac{d i}{d t}=-2 \mathrm{~A} / \mathrm{s}\) and \(L=5 \mathrm{H}\)
\(\therefore\) The emf developing across the coil,
\(
e=-L \frac{d i}{d t}=-5 \times(-2)=10 \mathrm{~V}
\)
In a coil of self-inductance 0.5 H, the current varies at a constant rate from zero to 10 A in 2 s. The emf generated in the coil is
(c) We have, \(\frac{\Delta i}{\Delta t}=\frac{10}{2}=5 \mathrm{~V}\) emf, \(e=L \frac{\Delta i}{\Delta t}=0.5 \times 5=2.5 \mathrm{~V}\)
Self-inductance of a coil is 50 mH. A current of 1 A passing through the coil reduces to zero at steady rate in 0.1 s, the self-induced emf is
(b) Here, \(L=50 \times 10^{-3} \mathrm{H}\) and \(\frac{d i}{d t}=\frac{(1-0)}{0.1}=10\)
\(\mathrm{emf},|e|=\frac{L \cdot d I}{d t}=50 \times 10^{-3} \times 10=50 \times 10^{-2}=0.5 \mathrm{~V}\)
The self-inductance of a long solenoid cannot be increased by
(d) The self-inductance of a long solenoid is given by
\(
L=\frac{\mu_r \mu_0 n^2 A}{l}
\)
It is clear that, the self-inductance of a long solenoid does not depend upon the current flowing through it.
The self-inductance of a coil is \(L\). Keeping the length and area same, the number of turns in the coil is increased to four times. The self-inductance of the coil will now be
(d) Self-inductance of coil is directly proportional to square of number of turns in the coil, i.e.
\(
\begin{array}{ll}
& L \propto N^2 \\
\therefore & \frac{L_1}{L_2}=\frac{N_1^2}{N_2^2} \\
\therefore & \frac{L}{L_2}=\frac{N^2}{(4 N)^2} \\
\therefore & L_2=16 L
\end{array}
\)
In circular coil, when number of turns is doubled and resistance becomes \((1 / 4)\) th of initial, then inductance becomes
(a) \(L \propto N^2\)
So, when number of turns is doubled, then \(L^{\prime}\) will become 4 times of \(L\), i.e. \(L^{\prime}=4 L\) or 4 times.
The self-inductance of solenoid of length \(L\), area of cross-section \(S\) and having \(N\) turns is
(a)
\(
\begin{aligned}
&\text { The self-inductance } L_{\text {ind }} \text { of a solenoid is given by: }\\
&L_{\text {ind }}=\frac{\mu_0 N^2 S}{L}
\end{aligned}
\)
A solenoid has 2000 turns would over a length of 0.30 m. The area of its cross-section is \(1.2 \times 10^{-3} \mathrm{~m}^2\). If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is
\(
\begin{aligned}
& \text { (d) Induced emf, } e=L \frac{d i}{d t}=\frac{\mu_0 N^2 S}{l} \frac{d i}{d t} \\
& \quad=\frac{4 \pi \times 10^{-7} \times(2000)^2 \times 1.2 \times 10^{-3}}{0.30} \times\left(\frac{2-(-2)}{0.25}\right) \\
& \quad=32.1 \times 10^{-2} \mathrm{~V}
\end{aligned}
\)
A 50 mH coil carries a current of 2 A. The energy stored in joules is
\(
\begin{aligned}
&\text { (b) Energy stored in the coil, }\\
&E=\frac{1}{2} L i^2=\frac{1}{2} \times 50 \times 10^{-3} \times 4=0.1 \mathrm{~J}
\end{aligned}
\)
In an inductor of inductance \(L=100 \mathrm{mH}\), a current of \(i=10 \mathrm{~A}\) is flowing. The energy stored in the inductor is
\(
\text { (a) } E=\frac{1}{2} L i^2=\frac{1}{2}\left(100 \times 10^{-3}\right)(10)^2=5 \mathrm{~J}
\)
Two pure inductors each of self-inductance \(L\) are connected in parallel but are well separated from each other. The total inductance is
(c) Equivalent inductance, \(\frac{1}{L_{\mathrm{eq}}}=\frac{1}{L}+\frac{1}{L}\)
The total inductance, \(L_{\mathrm{eq}}=\frac{L}{2}\)
Two coils \(X\) and \(Y\) are placed in a circuit such that when the current changes by 2 A in coil \(X\). The magnetic flux changes by 0.4 Wb in \(Y\). The value of mutual inductance of the coils is
(a) We have, \(M=\frac{\Delta \phi_2}{\Delta i_1} \Rightarrow M=\frac{0.4}{2}\)
\(\therefore\) Mutual inductance, \(M=0.2 \mathrm{H}\)
Two coils have a mutual inductance of 0.005 H . The current changes in the first coil according to equation \(i=i_0 \sin \omega t\), where \(i_0=10 \mathrm{~A}\) and \(\omega=100 \pi \mathrm{rads}^{-1}\). The maximum value of emf in the second coil is (in volt)
\(
\begin{aligned}
&\text { (b) Value of induced emf in coil, }\\
&\begin{aligned}
e & =M \frac{d i}{d t}=0.005 \times \frac{d}{d t}\left(i_0 \sin \omega t\right) \\
& =0.005 \times i_0 \omega \cos \omega t \quad\left(\text { for } e_{\max }, \cos \omega t=1\right) \\
e_{\max } & =0.005 \times 10 \times 100 \pi=5 \pi
\end{aligned}
\end{aligned}
\)
The mutual inductance between a primary and secondary circuits is 0.5 H . The resistance of the primary and the secondary circuits are \(20 \Omega\) and \(5 \Omega\), respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of
(a) emf, \(e_2=M\left(\frac{d i_1}{d t}\right)=i_2 R_2\)
\(\therefore \quad\) Current rate, \(\frac{d i_1}{d t}=\frac{i_2 R_2}{M}=\frac{(0.4)(5)}{0.5}=4 \mathrm{As}^{-1}\)
The coefficient of mutual induction between two circuits is equal to the emf produced in one circuit, when the current in the second circuit is
(c) The induced emf \(\mathscr{E}\) in a circuit due to mutual induction is given by \(\mathscr{E}=-M \frac{d I}{d t}\).
Here, \(\boldsymbol{M}\) is the coefficient of mutual induction and \(\frac{d I}{d t}\) is the rate of change of current.
We are looking for the condition where \(\mathscr{E}=\boldsymbol{M}\).
We know \(\mathscr{E}=-M \frac{d I}{d t}\).
So, \(M=-M \frac{d I}{d t}\).
Divide both sides by \(-M:-1=\frac{d I}{d t}\).
The magnitude of the rate of change of current is \(\left|\frac{d I}{d t}\right|=1 \mathrm{As}^{-1}\).
The coefficient of mutual induction between two circuits is equal to the emf produced in one circuit when the current in the second circuit is changed at the rate of \(1 \mathrm{As}^{-1}\).
Two circuits have coefficient of mutual induction of 0.09 H. Average emf induced in the secondary by a change of current from 0 to 20 A in 0.006 s in the primary will be
\(
\text { (d) Average emf, } e=M \frac{d i}{d t}=0.09 \times \frac{20}{0.006}=300 \mathrm{~V}
\)
A current is varying at the rate of \(3 \mathrm{~A} / \mathrm{s}\) in a coil generates an emf of 8 mV in a nearby coil. The mutual inductance of the two coils is
(a) As, \(|e|=M \frac{d i}{d t} \Rightarrow 8 \times 10^{-3}=M \times 3\)
Mutual inductance, \(M=2.66 \mathrm{mH}\)
A solenoid is placed inside another solenoid, the length of both being equal carrying same magnitude of current. The parameters like radius and number of turns are in the ratio \(1: 2\) for the two solenoids. The mutual inductance on each other would be
(a) We have, \(M=\frac{\mu_0 N_1 N_2 S}{L}\)
Here, \(S\) is area of inner solenoid. So, \(N_1, N_2\) and \(S\) are same for both solenoids.
\(
\Rightarrow \quad M_{12}=M_{21}
\)
Two coils of self-inductances \(L_1\) and \(L_2\) are placed closer to each other, so that total flux in one coil is completely linked with other. If \(M\) is mutual inductance between them, then
\(
\begin{aligned}
&\text { (c) Mutual inductance between two coils, }\\
&\begin{array}{ll}
& M=-\frac{e_2}{\left(d i_1 / d t\right)}=-\frac{e_1}{\left(d i_2 / d t\right)} \\
\text { Also, } & e_1=-L_1 \frac{d i_1}{d t} \text { and } e_2=-L_2 \frac{d i_2}{d t} \\
\therefore & M^2=\frac{e_{f_2}}{\left(\frac{d i_1}{d t}\right)\left(\frac{d i_2}{d t}\right)}=L_1 L_2 \\
\Rightarrow & M=\sqrt{L_1 L_2}
\end{array}
\end{aligned}
\)
Two coils of self-inductances 2 mH and 8 mH are placed, so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is
\(
\begin{aligned}
&\text { (a) Mutual inductance between coils is }\\
&\begin{aligned}
M & =K \sqrt{L_1 L_2} \\
M & =1 \sqrt{2 \times 10^{-3} \times 8 \times 10^{-3}} \\
& =4 \times 10^{-3}=4 \mathrm{mH}
\end{aligned} \quad(\because K=1)
\end{aligned}
\)
An ideal coil of 10 H is joined in series with a resistance of \(5 \Omega\) and a battery of 5 V. After 2 s of joining, the current flowing in ampere in the circuit will be
(b) Growth of current in the circuit is given by
\(
i=i_0\left(1-e^{-R t / L}\right)
\)
where, \(i_0\) is peak value of current and \(i_0=\frac{5}{5}=1 \mathrm{~A}\)
\(
\therefore \quad i=1\left(1-e^{-5 \times 2 / 10}\right)=\left(1-e^{-1}\right) \mathrm{A}
\)
An \(L-R\) circuit has a cell of emf \(E\), which is switched ON at time \(t=0\). The current in the circuit after a long time will be
(b) In case of growth of current in an \(L-R\) circuit, the current in the circuit grows exponentially with time to the maximum value, \(i_0=E / R\).
During current growth in an \(L-R\) circuit, the time constant is the time in which the magnitude of current becomes
(c) Time interval, during which the current in an inductive circuit rises to \(63 \%\) of its maximum value is defined as time constant, i.e. \(I=0.63 I_0\).
An \(L-R\) circuit with a battery is connected at \(t=0\). Which of the following quantities is not zero just after the connection?
(d) At \(t=0\), the inductor prevents an instantaneous rise in current. Therefore, the current \(I\) in the circuit is zero at \(t=0\).
Magnetic field energy \(U=\frac{1}{2} L I^2\).
Since \(I=0\) at \(t=0\), the magnetic field energy \(U\) is also zero.
Power delivered by the battery \(P=\operatorname{Emf} \times I\).
Since \(I=0\) at \(t=0\), the power delivered by the battery \(P\) is also zero.
The induced emf \(\mathscr{E}_{\text {ind }}=-L \frac{d I}{d t}\).
Although \(I=0\) at \(t=0\), the current is changing from zero, meaning \(\frac{d I}{d t} \neq 0\). Therefore, the induced emf \(\mathscr{E}_{\text {ind }}\) is not zero at \(t=0\).
Eddy currents are produced when
(a) A metal is kept in a varying magnetic field.
Explanation: Eddy currents are induced in a conductor when the magnetic field around it changes, creating a circulating current within the metal.
Why other options are incorrect:
(b) a metal is kept in a steady magnetic field:
A steady magnetic field does not cause any change in magnetic flux, therefore no eddy currents are produced.
(c) a circular coil is placed in a magnetic field:
While a circular coil in a magnetic field can experience a magnetic force, eddy currents only occur if the magnetic field is changing.
(d) through a circular coil current is passed:
Passing current through a coil creates a magnetic field around it, but this doesn’t necessarily induce eddy currents within the coil itself unless the current is changing.
Which of the following is not an application of eddy currents?
(d) X-ray crystallography is not an application of eddy currents.
Explanation:
Eddy currents are induced currents that occur in a conductor when a changing magnetic field passes through it. X-ray crystallography, on the other hand, is a technique that uses the diffraction of X-rays to determine the crystal structure of a material. This process is based on the interaction of Xrays with electrons, not with magnetic fields, which is the basis for eddy currents.
You cannot copy content of this page