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A toroid of \(n\) turns, mean radius \(R\) and cross-sectional radius \(a\) carries current \(I\). It is placed on a horizontal table taken as \(x-y\) plane. Its magnetic moment \(\mathbf{m}\)
(c) Toroid: A toroid can be considered as a ring shaped closed solenoid. Hence it is like an endless cylindrical solenoid.
The magnetic field is only confined inside the body of a toroid in the form of concentric magnetic lines of force. For any point inside the empty space surrounded by toroid and outside the toroid, the magnetic field \(\mathbf{B}\) is zero because the net current enclosed in these spaces is zero. Thus, the magnetic moment of toroid is zero.
Consider the two idealized systems: (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length \(L \gg R\), radius of cross-section. In (i) \(\mathbf{E}\) is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below:
(b) The electrostatic field lines, do not form a continuous closed path (this follows from the conservative nature of electric field) while the magnetic field lines form the closed paths.
According to Gauss’s law of electrostatic field \(\oint_s E . d s=\frac{q}{\varepsilon_0}\). so it does not contradict for electrostatic field as the electric field lines do not form continuous path. According to Gauss’s law of magnetic field \(\oint_s B . d s=0\). It contradicts for magnetic field, because there is magnetic field inside the solenoid and no field outside the solenoid marrying current, but the magnetic field lines from the closed paths. This implies that number of magnetic field lines entering the Gaussian surface is equal to the number of magnetic field lines leaving it.
A paramagnetic sample shows a net magnetisation of \(8 \mathrm{Am}^{-1}\) when placed in an external magnetic field of 0.6 T at a temperature of 4 K . When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K , the magnetisation will be
(b) \(I(\text { magnetization }) \propto \frac{B(\text { magnetic field induction })}{t(\text { temperature in kelvin })} \Rightarrow \frac{I_2}{I_1}=\frac{B_2}{B_1} \times \frac{t_1}{t_2}\)
Let us suppose, here \(I_1=8 \mathrm{Am}^{-1}\)
\(
B_1=0.6 \mathrm{~T}, \quad t_1=4 \mathrm{~K}
\)
\(
\begin{aligned}
B_2 & =0.2 \mathrm{~T}, \quad t_2=16 \mathrm{~K} \\
I_2 & =?
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{0.2}{0.6} \times \frac{4}{16}=\frac{I_2}{8} \\
& \Rightarrow \quad I_2=8 \times \frac{1}{12}=\frac{2}{3} \mathrm{Am}^{-1}
\end{aligned}
\)
The primary origin(s) of magnetism lies in
(a, d) The primary origin of magnetism lies in the fact that the electrons are revolving and spinning about the nucleus of an atom, and we know that an moving charge carries current along with it. We meant this current here as atomic current and which is responsible to produce an orbital magnetic moment. This atomic current gives rise to magnetism. The revolving and spinning about nucleus of an atom is called intrinsic spin of electron, which gives rise to spin magnetic moment. So, total magnetic moment is the sum of orbital magnetic moment and spin magnetic moment.
A long solenoid has 1000 turns per metre and carries a current of 1 A . It has a soft iron core of \(\mu_r=1000\). The core is heated beyond the Curie temperature, \(T_c\).
(a, d) The magnetic field intensity \(\mathrm{H}=\mathrm{nI}\), where \(\mathrm{n}=\) number of turns per metre of a solenoid and \(\mathrm{I}=\) current and \(\mathrm{B}=\mu_0 \mu_{\mathrm{r}} \mathrm{I}\).
Also, at normal temperature, a solenoid behaves as a ferromagnetic substance and at the temperature beyond the Curie temperature, it behaves as a paramagnetic substance.
\(\mathrm{n}=1000\) turns per metre, \(\mu_{\mathrm{r}}=1000\)
\(\mathrm{H}=\mathrm{nI}=1000 \times 1=1000 \mathrm{Amp}\). So \(\mathrm{H}\) is constant verifies the answer a
\(
\mathrm{B}=\mu_0 \mu_{\mathrm{r}} \mathrm{nl}=\left(\mu_0 \mathrm{nl}\right) \mu_{\mathrm{r}}=\mathrm{K} \mu_{\mathrm{r}}(\mathrm{~K}=\text { constant })
\)
\(
\text { So, } \mathrm{B} \propto \mu_r
\)
but there is a large decrease in the susceptibility of the core on heating it beyond critical temperature, hence magnetic field will decrease drastically. Now, for magnetisation in the core, when temperature of the iron core of a solenoid is raised beyond Curie temperature, then it behaves as a paramagnetic material, where
\(
\begin{aligned}
& \left(\chi_m\right)_{\text {Fero }} \approx 10^3 \\
\text { and } & \left(\chi_m\right)_{\text {Para }} \approx 10^{-5} \\
\Rightarrow & \frac{\left(\chi_m\right)_{\text {Fero }}}{\left(\chi_m\right)_{\text {Para }}}=\frac{10^3}{10^{-5}}=10^8
\end{aligned}
\)
Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to
(a, c, d) Electrostatic shielding is the phenomenon to block the effects of an electric field. The conducting shell can block the effects of an external field on its internal content or the effect of an internal field on the outside environment. For protecting a sensitive equipment from the external magnetic field it should be placed inside an iron cane (magnetic shielding). Magnetostatic shielding is done by using an enclosure made of a high permeability magnetic material to prevent a static magnetic field outside the enclosure from reaching objects inside it or to confine a magnetic field within the enclosure.
A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the axis of the loop is in
(a) Points lying on the axis of a magnet are called end-on points. In our case, the point on the axis of the loop (on replacing the circular loop with an equivalent magnetic dipole) lies on the axis of the magnetic dipole or on the end-on position. If P was the point on the axis of the loop, then it is clear from the figure that P lies on the end-on position of the equivalent magnetic dipole.
Note: A point on the axis of a current-carrying circular loop, when replaced by an equivalent magnetic dipole, is in the end-on position. This is because the axis of the loop is aligned with the axis of the equivalent magnetic dipole.
Explanation:
Magnetic Dipole:
A current-carrying circular loop acts as a magnetic dipole, with a magnetic moment determined by the current and the loop’s area.
End-on Position:
When a point is on the axis of the magnetic dipole, it’s considered to be in the end-on position according to physics experts.
Broadside-on Position:
The broadside-on position is a point perpendicular to the axis of the dipole.
Current Loop and Dipole:
The magnetic field produced by a current loop at a distance is similar to the magnetic field produced by a magnetic dipole at a corresponding distance. The axis of the dipole is aligned with the axis of the loop.
A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in
(b) A point on a circular loop carrying a current, when considered as an equivalent magnetic dipole, is in a broadside-on position.
Explanation: A broadside-on position refers to a point on the perpendicular bisector of the dipole, which is analogous to a point on the plane of the circular loop when it’s treated as a magnetic dipole. The magnetic field at this point is perpendicular to the loop’s plane, just like the magnetic field of a dipole is perpendicular to the line connecting the poles at its broadside.
When a current in a circular loop is equivalently replaced by a magnetic dipole,
(c) When a circular loop carrying current is replaced by a magnetic dipole, the product of the pole strength ( \(m\) ) and the distance between the poles ( \(d\) ), which is the magnetic dipole moment, remains constant. While the individual values of \(m\) and \(d\) can vary, their product must be consistent to maintain the same magnetic field.
Explanation:
Magnetic dipole moment:
The magnetic dipole moment ( \(M\) ) of a current loop is given by \(M=I A\), where \(I\) is the current and A is the area of the loop.
Equivalent magnetic dipole:
A magnetic dipole consists of two equal and opposite magnetic poles separated by a distance. The magnetic dipole moment of this dipole is \(m d\), where \(m\) is the pole strength and \(d\) is the distance between the poles.
Equivalence:
The key is that the external magnetic field produced by both the current loop and the equivalent magnetic dipole must be identical. This means their magnetic dipole moments must be the same.
Therefore, \(I A=m d\). While you can change the values of ‘ \(m\) ‘ and ‘ \(d\) ‘, their product must remain constant to match the magnetic field of the current loop. Thus, the product md is fixed.
Let \(r\) be the distance of a point on the axis of a bar magnet from its centre. The magnetic field at such a point is proportional to
(d) Magnetic field \(B\) due to a bar magnet of magnetic moment \(M\) at distance \(r\) of the point on the axis of the magnet from its centre is given by
\(
B=\frac{u_0}{4 \pi} \frac{2 M r}{\left(r^2-l^2\right)^2}
\)
So, from the above formula, it can be easily seen that \(B \propto \frac{r}{\left(r^2-l^2\right)^2}\)
Let \(r\) be the distance of a point on the axis of a magnetic dipole from its centre. The magnetic field at such a point is proportional to
(c) As, we know that, the magnetic field at a point due to the dipole is given by \(B=\frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3} \sqrt{1+3 \cos ^2 \theta}\)
where, \(M\) is is magnetic dipole moment.
\(
\Rightarrow \quad B \propto \frac{1}{r^3}
\)
Two short magnets of equal dipole moments \(M\) are fastened perpendicularly at their centres (figure below). The magnitude of the magnetic field at a distance \(d\) from the centre on the bisector of the right angle is
(c) These short magnets can be considered magnetic dipoles with dipole moment M each. For a point P, at distance d from the dipoles (as shown in the diagram below), the radial strength of the field due to each dipole
\(
\begin{aligned}
B^{\prime} & =\mu_0 2 M \cos 45^{\circ} / 4 \pi d^3 \\
& =\mu_0 2 M /\left(4 \sqrt{ } 2 \pi d^3\right) \\
& =\mu_0 M /\left(2 \sqrt{ } 2 \pi d^3\right)
\end{aligned}
\)
The strength of the component perpendicular to \(O P\) is \(B^{\prime \prime} . B^{\prime \prime}\) is equal in magnitude due to each dipole but opposite in direction. So they cancel each other. Thus only radial components remain in the same direction. Hence the net field at \(P\)
\(
\begin{aligned}
B & =B^{\prime}+B^{\prime}=2 B^{\prime} \\
& =2 \left(\mu_0 M / 2 \sqrt{ } 2 \pi d^3\right) \\
& =\mu_0 M /\left(\sqrt{ } 2 \pi d^3\right) \\
& =\mu_0 \sqrt{ } 2 M / 2 \pi d^3 \\
& =\mu_0 2 \sqrt{ } 2 M / 4 \pi d^3
\end{aligned}
\)
(multiplying numerator and denominator by 2 )
Magnetic meridian is
(d) a vertical plane.
Explanation: A magnetic meridian is an imaginary vertical plane that passes through the magnetic north and south poles of the Earth, essentially defining the direction a compass needle points at a given location.
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It
(d) Will stay in any position.
Explanation: At the magnetic pole the direction of the magnetic field is vertical. But the needle can move only in the horizontal plane. The horizontal component of the magnetic field here is zero. Hence there will be no effect of the magnetic field on the compass needle and it will stay in any position.
A dip circle is taken to geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay
(d) in any direction it is released.
Explanation: At the Geomagnetic equator, the direction of the magnetic field is horizontal in the magnetic meridian. So the component of the magnetic field in a plane perpendicular to the magnetic meridian is zero. So the needle of the dip circle that is free to move in this perpendicular plane will stay in any direction it is released.
Which of the following four graphs may best represent the current-deflection relation in a tangent galvanometer?
(c) In a tangent galvanometer the relation between the current and deflection is given as,
\(
\mathrm{i}=\mathrm{K} \times \tan \theta
\)
Where K is a constant.
For very small \(\theta, \tan \theta\) varies nearly linearly. As \(\theta\) approaches \(\pi / 2\), even a small increase in \(\theta\) increases \(\tan \theta\) by a great amount. And near \(\pi / 2 \tan \theta\) tends to infinity.
So in the tangent galvanometer, the deflection-current graph will be a straight line initially, and with the increase in deflection, the graph will curve and become asymptotic to the current axis near \(\pi / 2\). Only graph (c) fulfills the condition.
A tangent galvanometer is connected directly to an ideal battery. If the number of turns in the coil is doubled, the deflection will
(c) The magnetic field created by the coil at the center is
\(
B=\mu_0 i n / 2 r
\)
When the number of turns is kept double, i.e. \(=2 n\), the resistance of the coil is also doubled. It results in reducing the current to half i.e. \(\mathrm{i} / 2\). Now the magnetic field by the coil is \(\mu_o(i / 2)(2 n) / 2 r=\mu_o i n / 2 r=B\).
Thus the magnetic field remains unchanged and so will be the deflection.
If the current is doubled, the deflection is also doubled in
(b) In a tangent galvanometer the current is proportional to the tangent of the deflection. So doubling the current will not double the deflection.
In a moving coil galvanometer, the current is directly proportional to the deflection. Hence when we double the current the deflection will also get doubled. Option (b) is correct.
A very long bar magnet is placed with its north pole coinciding with the centre of a circular loop carrying an electric current \(i\). The magnetic field due to the magnet at a point on the periphery of the wire is \(B\). The radius of the loop is \(a\). The force on the wire is
In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current \(i\). So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring.
Let idl be the current element, \(B\) be the magnetic field and \(dF\) be the force on the current element \(idl\).
\(
\begin{aligned}
&\begin{aligned}
& d F=B i d l \Rightarrow F=\int_0^{2 \pi a} B i d l \\
& \Rightarrow F=2 \pi a i B
\end{aligned}\\
&\text { Thus, the force acting on the wire is } 2 \pi a i B \text { and it is perpendicular to the plane of the wire. }
\end{aligned}
\)
Pick the correct options.
(a, b) Explanation: Magnetic field is produced by moving charges only. Even in permanent magnets, the magnetic properties come from the moving charges at atomic levels. (a) is correct.
The magnetic poles are mathematical assumptions to study the magnetic effects. And we call a positive magnetic charge the North Pole and a negative magnetic charge the south pole. Option (b) is also correct.
(c) is not correct because a north pole is equivalent to an anticlockwise current and a south pole is equivalent to a clockwise current.
Option (d) is also not correct because a bar magnet is not equivalent to a long, straight current because the magnetic fields produced by them are different.
A horizontal circular loop carries a current that looks clockwise when viewed from above. It is replaced by an equivalent magnetic dipole consisting of a south pole \(S\) and a north pole \(N\).
(b, d) If a circular loop is replaced by an equivalent magnetic dipole, then the SN line should be perpendicular to the plane of the loop. It is because the north pole is towards the side from which the current in the loop is anticlockwise, and the south pole is to that side from which the current in the loop is viewed clockwise. Hence only the options (b) and (d) are correct.
Consider a magnetic dipole kept in the north-to-south direction. Let \(P_1, P_2, Q_1, Q_2\) be four points at the same distance from the dipole towards north, south, east and west of the dipole respectively. The directions of the magnetic field due to the dipole are the same at
(a, b) The points \(P_1\) and \(P_2\) are equidistant from the dipole on both sides of its axis. At \(\mathrm{P}_1\), the direction of the field will be from the dipole towards the north pole. Similarly at \(\mathrm{P}_2\), the direction of the field will also be towards the north and the south end of the dipole. Option (a) is correct.
Q1 and Q2 are symmetrical points on both sides of the perpendicular bisector of the dipole. Hence the directions of the magnetic field will also be the same at these points. The direction will be from north to south. Hence the option (b) is also correct.
The directions of the field at points in options (c) and (d) are opposite to each other. Hence not correct.
Consider the situation of the previous problem. The directions of the magnetic field due to the dipole are opposite at
(c, d) As we can see in the diagram below,
At points \(\mathrm{P}_1\) and \(\mathrm{Q}_1\) the directions of the magnetic field are opposite. Also at points \(\mathrm{P}_2\) and Q2 the directions are opposite. Hence only options (c) and (d) are correct.
To measure the magnetic moment of a bar magnet, one may use
(b, c, d) To measure the magnetic moment of a bar magnet, we can measure \(\mathrm{M} / \mathrm{B}_{\mathrm{H}}\) using a deflection magnetometer or \(\mathrm{MB}_{\mathrm{H}}\) using an oscillation magnetometer if we know \(\mathrm{B}_{\mathrm{H}}\). Hence options (b), (c) and (d) are correct not option (a).
Magnetic length is
(a) Less than the geometric length. The magnetic poles of a magnet are not located at the exact physical ends of the magnet, but rather slightly inside from the ends. The geometric length is the total length of the magnet, while the magnetic length is the distance between the two poles. Therefore, the magnetic length is always shorter.
Key points:
Geometric length: The physical length of the magnet, measured from end to end.
Magnetic length: The distance between the magnetic north and south poles.
Poles are inside: The poles are not located at the very ends of the magnet, but slightly within.
Magnetic lines of force due to a bar magnet do not intersect because
(a) Magnetic lines of force due to a bar magnet do not intersect because a point on magnetic lines always has a single net magnetic field. If they intersect then it means, there are two directions of magnetic field intensity at that point, which is impossible.
The unit of pole strength is
(a) Pole strength: When another magnet of opposite polarity is brought near a pole, the amount of force a pole of magnet exerts is called pole strength.
In SI units, there are two possible units for magnetic pole strength
It can be defined as the force per unit H field, in which case the SI unit is the weber (Wb).
If you define it as force per unit B field, then the SI unit is the amperemeter (A-m).
A bar magnet of magnetic moment \(M_1\) is axially cut into two equal parts. If these two pieces are arranged perpendicular to each other, the resultant magnetic moment is \(M_2\). Then, the value of \(M_1 / M_2\) is
(d) Initial magnetic moment \(=M_1\)
When cut in two parts, magnetic moment of each part
\(
M^{\prime}=\frac{M_1}{2}
\)
When these pieces are placed perpendicular, effective magnetic moment,
\(
\begin{aligned}
& M_2=\sqrt{2} M^{\prime}=\sqrt{2} \times \frac{M_1}{2} \\
& \frac{M_1}{M_2}=\frac{2}{\sqrt{2}}=\sqrt{2}
\end{aligned}
\)
At a point on the right bisector of a magnetic dipole, the magnetic
(b) The magnetic potential is given by \(V=\frac{\mu_0}{4 \pi} \frac{M \cos (\theta)}{r^2}\). On the right bisector, \(\theta=90^{\circ}\), so \(\cos \left(90^{\circ}\right)=0\).
Therefore, \(V=\frac{\mu_0}{4 \pi} \frac{M \cdot 0}{r^2}=0\).
The magnetic potential is zero at all points on the right bisector, and the field is perpendicular to the axis of the dipole.
The ratio of the magnetic fields due to small bar magnet in end on position to broadside on position is (at equal distance from the magnet)
(d) For the axial position (end position):
\(
B_{\text {axial }}=\frac{\mu_0}{4 \pi} \cdot \frac{2 m}{d^3}
\)
For the equatorial position (broad side position):
\(
B_{\text {equatorial }}=\frac{\mu_0}{4 \pi} \cdot \frac{m}{d^3}
\)
\(
\frac{B_{\text {axial }}}{B_{\text {equatorial }}}=\frac{2 m}{m}=2
\)
Two solenoids acting as short bar magnets \(P\) and \(Q\) are arranged such that their centres are on the \(X\)-axis and are separated by a large distance. The magnetic axes of \(P\) and \(Q\) are along \(X\) and \(Y\)-axes, respectively. At a point \(R\), mid-way between their centres, if \(B\) is the magnitude of induction due to \(Q\), then the magnitude of total induction at \(R\) due to the both magnets is
(b) Since, for magnet \(P\), axis lies along \(X\)-axis and for magnet \(Q\), axis is along \(Y\)-axis. The point \(R\) is along axial line w.r.t. magnet \(P\) and is along equatorial line w.r.t. magnet \(Q\).
Hence, magnetic field due to magnet \(Q\),
\(
B_Q=\frac{\mu_0}{4 \pi} \frac{M}{x^3}=B \quad[R \text { at equatorial point }] \ldots \text { (i) }
\)
\(
\begin{aligned}
&\text { Magnetic field due to magnet } P \text {, }\\
&B_P=\frac{\mu_0}{4 \pi} \frac{2 M}{x^3}=2 B \text { [ } R \text { at axial point] } \ldots \text {.(ii) }
\end{aligned}
\)
As, at point \(R\) magnetic field due to \(P\) and \(Q\) magnet are perpendicular to each other and \(B_R=\) net magnetic field at \(R\) due to magnet \(P\) and \(Q\), i.e.
\(
B_R=\sqrt{B_P^2+B_Q^2}=\sqrt{B^2+(2 B)^2}=\sqrt{5} B
\)
The intensity of magnetic field due to an isolated pole of strength \(m\) at a point distant \(r\) from it will be proportional to
The magnetic field intensity due to an isolated pole is given by \(m_p=\frac{\boldsymbol{m}}{4 \pi \mu_0 \boldsymbol{r}^2}\).
\(\mu_0\) is the permeability of free space.
\(
m_p \propto \frac{m}{r^2}
\)
The magnetic field intensity is proportional to \(\frac{m}{r^2}\).
A particle of charge \(q\) and mass \(m\) moves in a circular orbit of radius \(r\) with angular speed \(\omega\). The ratio of the magnitude of its magnetic moment to that of its angular momentum is
(a) The relation between magnetic moment \(\left(\mu_l\right)\) and angular momentum \((L)\) is
\(
\mu_l=-\frac{q}{2 m} L \Rightarrow \frac{\mu_l}{L}=-\frac{q}{2 m}
\)
The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.
A bar magnet of magnetic moment \(\mathbf{M}\), is placed in a magnetic field of induction \(\mathbf{B}\). The torque exerted on it is
(a) When a bar magnet of magnetic moment \(\mathbf{M}\) is placed in a magnetic field of induction \(\mathbf{B}\), then net force on it is
\(
F_R=m B+(-m) B=0
\)
While torque, \(\tau=m B \times 2 l \sin \theta=M B \sin \theta \quad\) (as, \(M=2 \mathrm{ml}\) ) i.e.
\(
\tau=\mathbf{M} \times \mathbf{B}
\)
The couple acting on a magnet of length 10 cm and pole strength \(15 \mathrm{~A}-\mathrm{m}\), kept in a field of \(B=2 \times 10^{-5} \mathrm{~T}\), at an angle of \(30^{\circ}\) is
(a)
\(
\begin{aligned}
\tau & =M B \sin \theta=(m \times 2 l) \times 2 \times 10^{-5} \sin 30^{\circ} \\
& =15 \times 10 \times 10^{-2} \times 2 \times 10^{-5} \times \frac{1}{2} \\
& =1.5 \times 10^{-5} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)
A bar magnet is held at right angle to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it from this position. The angle of rotation is
(a) As, \(\tau=M B \sin \theta\)
where, \(\theta=90^{\circ} \Rightarrow \tau=M B\)
Given, \(\tau_2=\frac{1}{2} \tau_1 \Rightarrow M B \sin \theta=\frac{1}{2} M B\)
\(\Rightarrow \quad \sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}\)
\(\therefore\) Angle of rotation \(=90^{\circ}-30^{\circ}=60^{\circ}\)
If a bar magnet of magnetic moment \(M\) is freely suspended in a uniform magnetic field of strength \(B\), then the work done in rotating the magnet through an angle \(\theta\) is
(d) Work done is rotating the magnet through an angle \(\theta\) from initial position (i.e. \(\theta_1=0^{\circ}\) ) is given by
\(
W=M B\left(\cos \theta_1-\cos \theta\right)=M B\left(\cos 0^{\circ}-\cos \theta\right)=M B(1-\cos \theta)
\)
The effect due to uniform magnetic field on a freely suspended magnetic needle is
(b) torque is present but no net force.
Explanation:
When a magnetic needle is placed in a uniform magnetic field, the equal and opposite forces exerted on its north and south poles cancel out, resulting in no net force. However, these forces create a torque, causing the needle to rotate until it aligns with the magnetic field.
Why other options are incorrect:
(a) Both torque and net force are present: As explained above, the net force is zero in a uniform magnetic field. While a torque is present, making (a) incorrect.
(c) Both torque and net force are absent: Since a torque is always present in this scenario, (c) is incorrect.
(d) net force is present but not torque: The equal and opposite forces create a net force which is zero, so (d) is incorrect.
The net magnetic flux through any closed surface kept in a magnetic field is
(a) According to Gauss’ theorem in magnetism, surface integral of magnetic field intensity over a surface (closed or open) is always zero,
i.e. \(\oint \mathbf{B} \cdot d \mathbf{A}=0\).
The earth’s magnetic field is
(d) The strength of earth’s magnetic field at the surface of the earth ranges from 0.25 G to 0.65 G.
Magnetic meridian is a
(d) vertical plane. It’s an imaginary plane that passes through the magnetic north and south poles, and it’s where a compass needle will align itself. While it can be described as a line along the north-south direction, it’s more accurately a vertical plane.
The angle between the magnetic meridian and geographical meridian is called
(b) angle of declination.
Explanation:
Angle of declination:
This is the angle between the magnetic meridian (where a compass needle points) and the geographical meridian (true north) at a given location on Earth.
Angle of dip:
This refers to the angle the Earth’s magnetic field lines make with the horizontal plane at a given location.
Magnetic moment:
This is a measure of the strength and direction of a magnet’s magnetic field.
Power of magnetic field:
This is not a standard term used to describe the relationship between the magnetic meridian and geographical meridian.
Angle of dip at the equator is
(a) At the equator, the Earth’s magnetic field is parallel to the surface. Therefore, the angle of dip is \(0^{\circ}\).
Earth’s magnetic field always has a horizontal component except at
(b) The Earth’s magnetic field always has a horizontal component except at the magnetic pole. At the magnetic poles, the magnetic field lines are vertical, resulting in a horizontal component of zero. At the magnetic equator, the horizontal component is at its maximum. The geographical north pole and the magnetic north pole are not exactly the same, but at both, the magnetic field is vertical.
If \(H=\frac{1}{\sqrt{3}} V\), then find angle of dip. (where symbols have their usual meanings)
(a) Magnetic dip or magnetic inclination is given by
\(
\tan \theta=\frac{V}{H} \dots(i)
\)
where, \(V\) and \(H\) are vertical and horizontal components of earth’s magnetic field, respectively.
\(
\text { Given, } H=\frac{1}{\sqrt{3}} V \Rightarrow \frac{V}{H}=\sqrt{3} \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\tan \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}
\)
Let \(V\) and \(H\) be the vertical and horizontal components of earth’s magnetic field at any point on earth. Near the north pole
(a) Near the Earth’s north pole, the vertical component of the Earth’s magnetic field is much larger than the horizontal component, making the correct answer (a) \(V \gg H\).
If a magnet is suspended at angle \(30^{\circ}\) to the magnetic meridian, then the dip needle makes angle of \(45^{\circ}\) with the horizontal. The real dip is
\(
\begin{aligned}
\text { (d) As, } \tan \delta^{\prime} & =\frac{\tan \delta}{\cos \theta}=\frac{\tan 45^{\circ}}{\cos 30^{\circ}} \\
\tan \delta^{\prime} & =\frac{1}{\sqrt{3} / 2}=\frac{2}{\sqrt{3}} \\
\Rightarrow \quad \delta^{\prime} & =\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)
\end{aligned}
\)
The dip at a place is \(\delta\). For measuring it, the axis of the dip needle is perpendicular to the magnetic meridian. If the axis of the dip needle makes an angle \(\theta\) with the magnetic meridian, then the apparent dip will be given by
(a)
\(
\begin{aligned}
\tan \delta & =\frac{V}{H} \text { and } \tan \delta_1=\frac{V}{H \cos \left(90^{\circ}-\theta\right)} \\
& =\frac{\tan \delta}{\cos \left(90^{\circ}-\theta\right)}=\tan \delta \operatorname{cosec} \theta
\end{aligned}
\)
At a neutral point
(d) None of the above is true.
Explanation: A neutral point is a location where the magnetic field due to a magnet cancels out the Earth’s magnetic field, resulting in a net magnetic field of zero.
Why the other options are incorrect:
(a) Field of magnet is zero:
The magnet still creates a magnetic field, but it’s counteracted by the Earth’s field at the neutral point.
(b) Field of earth is zero:
The Earth’s magnetic field is constant, it’s the magnet’s field that is being balanced at the neutral point.
(c) Field of magnet is perpendicular to field of earth:
While the fields are opposite in direction at a neutral point, they are not necessarily perpendicular. Perpendicularity would imply a right angle, which isn’t the condition for cancellation, just opposite direction and equal magnitude.
Tangent galvanometer measures
(b) A tangent galvanometer measures current.
Explanation: A tangent galvanometer uses the principle of tangent law to compare the magnetic field produced by the current to the Earth’s magnetic field, allowing for the measurement of current strength.
When a current is passed through the galvanometer coil, then a magnetic field \(B\) is produced at right angles to the plane of the coil, ie, at right angles to the horizontal component of earth’s magnetic field \(H\). Under the influence of two crossed magnetic fields \(B\) and \(H\), the magnetic needle of galvanometer
undergoes a deflection \(\theta\) which is given by the tangent law. Using tangent law, we can find a relation
\(I \propto \tan \theta\)
Which clearly indicates that tangent galvanometer is an instrument used for detection of electric current in a circuit.
Note A tangent galvanometer is most accurate when its deflection is \(45^{\circ}\)
Two tangent galvanometers \(A\) and \(B\) are identical except in their number of turns. They are connected in series. On passing a current through them, deflections of \(60^{\circ}\) and \(30^{\circ}\) are produced. The ratio of the number of turns in \(A\) and \(B\) is
(b) As, the current and the other factors are same for both the galvanometers, so
\(
N \propto \tan \theta \Rightarrow \frac{N_1}{N_2}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}} \text { or } \frac{N_1}{N_2}=\frac{\sqrt{3}}{1 / \sqrt{3}}=3: 1
\)
Vibration magnetometer is used for comparing
(d) All of these. It can be used to compare magnetic fields, the Earth’s field, and magnetic moments.
The time period of a freely suspended bar magnet in a field is 2 s. It is cut into two equal parts along its axis, then the time period is
\(
\begin{aligned}
&\text { (c) As, time period } T=2 \pi \sqrt{\frac{I}{M B}}\\
&\begin{array}{ll}
\because & T=2 \mathrm{~s}, I^{\prime}=\frac{I}{2}, M^{\prime}=\frac{M}{2} \\
\Rightarrow & T^{\prime}=T=2 \mathrm{~s}
\end{array}
\end{aligned}
\)
A bar magnet suspended freely in a uniform magnetic field is vibrating with a time period of 3 s. If the field strength is increased to 4 times of the earlier field strength, then the time period (in second) will be
\(
\text { (c) As, the time period, } T=2 \pi \sqrt{\frac{I}{M B}}
\)
\(
\begin{aligned}
& T \propto \frac{1}{\sqrt{B}} \Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}} \\
& \frac{T_2}{3}=\sqrt{\frac{1}{4}} \Rightarrow T_2=1.5 \mathrm{~s}
\end{aligned}
\)
An example of a diamagnetic substance is
(b) copper.
Explanation:
Diamagnetic substances: are materials that are weakly repelled by a magnetic field. They have no unpaired electrons and thus do not exhibit magnetism on their own. –
Copper: is a well-known example of a diamagnetic material.
Aluminium: is paramagnetic, meaning it is weakly attracted to a magnetic field.
Iron and nickel: are ferromagnetic, meaning they are strongly attracted to a magnetic field and can be magnetized.
The universal property of all substances is
(a) diamagnetism.
Explanation: Diamagnetism is the property inherent to all materials, meaning every substance exhibits some degree of diamagnetic behavior, where it is slightly repelled by a magnetic field.
Why other options are incorrect:
(b) Ferromagnetism:
This property is only present in a few specific materials like iron, nickel, and cobalt, where they are strongly attracted to a magnetic field and can retain their magnetization even after the external field is removed.
(c) Paramagnetism:
While more common than ferromagnetism, paramagnetism is still not universal. Paramagnetic materials are weakly attracted to a magnetic field due to unpaired electrons in their atoms.
(d) All of the above:
Since only diamagnetism is present in all substances, option (d) is incorrect.
Which magnetic materials have negative susceptibility?
(a) Diamagnetic materials.
Explanation:
Diamagnetic materials have a negative magnetic susceptibility because their magnetic dipoles align opposite to an applied magnetic field, meaning they are repelled by the field.
Why other options are incorrect:
(b) Paramagnetic materials:
Paramagnetic materials have a positive magnetic susceptibility because their dipole align partially in the same direction as the applied magnetic field, resulting in a net attraction to the field.
(c) Ferromagnetic materials:
Ferromagnetic materials have a very large positive susceptibility due to their strong tendency to align dipoles parallel to each other, even in the absence of an external field.
(d) All of these: Since only diamagnetic materials have a negative susceptibility, this option is incorrect.
Identify the paramagnetic substance.
(b) Aluminium.
Explanation:
Paramagnetic substances:
These are materials that are weakly attracted to a magnetic field, but lose their magnetization when the field is removed. Aluminium is classified as a paramagnetic material.
Other options:
Iron: Iron is actually a ferromagnetic substance, meaning it has a strong attraction to a magnetic field and retains its magnetization even after the field is removed.
Nickel: Like iron, nickel is also ferromagnetic.
Hydrogen: Hydrogen is considered a diamagnetic substance, meaning it is very weakly repelled by a magnetic field.
Which of the following is true?
(b) Paramagnetism is temperature dependent.
Explanation:
Diamagnetism:
This property is due to the paired electrons in an atom resisting an external magnetic field, and is essentially independent of temperature.
Paramagnetism:
This property arises from unpaired electrons which align with an external magnetic field, meaning their alignment increases with temperature, making the magnetic susceptibility dependent on temperature.
Why other options are incorrect:
(a) Diamagnetism is temperature dependent: This is false, as diamagnetism is temperature independent.
(c) Paramagnetism is temperature independent: This is false, as paramagnetism is temperature dependent.
(d) None of the above: This is incorrect because option (b) is true.
Magnetic permeability is maximum for
(c) ferromagnetic substances.
Explanation: Ferromagnetic materials have a significantly higher permeability compared to diamagnetic and paramagnetic substances, meaning they are strongly attracted to magnetic fields and can retain their magnetization even after the external field is removed.
Key points about different types of magnetic substances:
Diamagnetic substances:
These have a slightly lower permeability than vacuum, meaning they are weakly repelled by magnetic fields.
Paramagnetic substances:
These have a slightly higher permeability than vacuum, meaning they are weakly attracted to magnetic fields.
Ferromagnetic substances:
These have a much higher permeability than other materials, allowing them to be strong magnets.
The temperature at which a ferromagnetic material becomes paramagnetic is called a
(b) Curie temperature. The Curie temperature is the specific temperature at which a ferromagnetic material loses its strong magnetic properties and becomes paramagnetic.
Here’s why the other options are incorrect:
(a) neutral temperature:
This term generally refers to the environmental temperature range where the body maintains its core temperature with minimal effort, not a magnetic transition temperature.
(c) inversion temperature:
This refers to a temperature inversion in the atmosphere where temperature increases with altitude, not a magnetic transition.
(d) critical temperature:
This term is used in thermodynamics to describe the highest temperature at which a substance can exist as a liquid, regardless of pressure.
Substances in which the magnetic moment of a single atom is not zero, is known as
(c) paramagnetism.
Explanation: Paramagnetic materials are characterized by having a net magnetic moment in their atoms, meaning their individual atoms possess a magnetic dipole moment that is not cancelled out by other moments within the same atom. This results in a weak attraction to magnetic fields. Diamagnetism (a) is the opposite, where the magnetic moment of a single atom is zero due to paired electrons. Ferromagnetism (b) and ferrimagnetism (d) are types of magnetism where atoms align in a specific way, creating strong magnetic properties. However, the question specifically asks about the magnetic moment of a single atom, which is a characteristic of paramagnetism.
Liquid oxygen remains suspended between two pole faces of a magnet because it is
(b) Liquid oxygen remains suspended between two pole faces of a magnet because it is paramagnetic.
Explanation: Paramagnetic materials are attracted to magnetic fields due to the presence of unpaired electrons in their molecules. Liquid oxygen has unpaired electrons in its oxygen molecules, which allows it to be attracted to a strong magnetic field, causing it to be suspended between the poles.
The only property possessed by ferromagnetic substance is
(a) Some peculiar properties of ferromagnetic materials are commonly displayed by curve of \(\mathbf{B}\) against \(\mathbf{H}\), which is called \(B\) – \(H\) curve or hysteresis loop. Diamagnetic and paramagnetic substances do not show these properties.
The permanent magnet is made from which one of the following substances?
(c) If a magnet retains its attracting power for a long time it is said to be permanent, otherwise temporary. Permanent magnets are made of ferromagnetic substances.
The SI unit of magnetic permeability is
\(
\begin{aligned}
&\text { (c) Magnetic permeability, }\\
&\mu=\frac{B}{H}
\end{aligned}
\)
\(
\text { Its SI unit is } \frac{\mathrm{Wb} / \mathrm{m}^2}{\mathrm{~A} / \mathrm{m}}=\frac{\mathrm{Wb}}{\mathrm{~A}} \cdot \frac{1}{\mathrm{~m}}=\text { Henry } m^{-1}\left[\because \frac{\mathrm{~Wb}}{\mathrm{~A}}=\mathrm{H} \text { (Henry) }\right]
\)
The unit of magnetic susceptibility is
(d) The unit of magnetic susceptibility is dimensionless.
Explanation
Magnetic susceptibility ( \(\chi\) ) is defined as the ratio of magnetization ( M ) to the applied magnetic field strength \((\mathrm{H})\) :
\(
\chi=\frac{M}{H}
\)
Both magnetization \((\mathrm{M})\) and magnetic field strength \((\mathrm{H})\) are measured in units of Amperes per meter (A/m). Therefore, the ratio of M to H is dimensionless.
Which one of the following is a non-magnetic substance?
(d) Brass.
Explanation:
Brass is an alloy made of copper and zinc, neither of which are magnetic.
Therefore, brass is not magnetic.
Why other options are incorrect:
(a) Iron: Iron is a ferromagnetic material, meaning it is strongly attracted to magnets.
(b) Nickel: Nickel is also a ferromagnetic material, meaning it is highly magnetic.
(c) Cobalt: Cobalt is another ferromagnetic material, meaning it is attracted to magnets.
The relation between \(B, H\) and \(I\) in SI system is
(c)
\(
\begin{aligned}
&\text { The relation between } B, H \text {, and } I \text { in the } \mathrm{SI} \text { system is given by: }\\
&B=\mu_0(H+I)
\end{aligned}
\)
The magnetic susceptibility of a paramagnetic material at \(-73^{\circ} \mathrm{C}\) is 0.0075 . Find its value at \(-173^{\circ} \mathrm{C}\).
(b) Magnetic susceptibility,
\(
\begin{aligned}
& \chi_{m_1}=0.0075, T_1=-73^{\circ} \mathrm{C}=(-73+273) \mathrm{K}=200 \mathrm{~K} \\
& \chi_{m_2}=?, T_2=-173^{\circ} \mathrm{C}=(-173+273) \mathrm{K}=100 \mathrm{~K}
\end{aligned}
\)
According to Curie’s law, \(\chi_m \propto \frac{1}{T}\)
\(\therefore\) Ratio of magnetic susceptibilities, \(\frac{\chi_{m_2}}{\chi_{m_1}}=\frac{T_1}{T_2}=\frac{200}{100}=2\)
\(\Rightarrow \quad \chi_{m_2}=2 \chi_{m_1}=2 \times 0.0075=0.015\)
A paramagnetic material is placed in a magnetic field.
(A) If the magnetic field is increased, the magnetization Consider the following statements: is increased.
(B) If the temperature is increased, the magnetization is increased.
(b) If the magnetic field is increased, magnetisation of paramagnetic material placed in magnetic field is also increases. Hence, option (a) is correct.
Magnetization (\(I\)) is given by,
\(
\vec{I}=\frac{\vec{M}}{V}
\)
As the temperature is increased, magnetic moments of paramagnetic material becomes more randomly aligned due to incresed thermal motion. This leads decrease in the magnetization \(I\).
A paramagnetic material is kept in a magnetic field. The field is increased till the magnetization becomes constant. If the temperature is now decreased, the magnetization
(c) If a paramagnetic material is subjected to an increasing magnetic field until its magnetization becomes constant (saturation point), and then the temperature is decreased, the magnetization will remain constant.
Explanation:
At the saturation point, all the magnetic dipoles within the paramagnetic material are aligned with the external magnetic field. Decreasing the temperature further will reduce thermal vibrations, but it won’t cause any additional alignment because the dipoles are already fully aligned.
A ferromagnetic material is placed in an external magnetic field. The magnetic domains
(c) When a ferromagnetic material is placed in an external magnetic field, the magnetic domains may increase or decrease in size. The domains aligned with the field will grow, while those opposed to the field will shrink. This happens because the external field influences the orientation of magnetic dipoles within the domains, causing them to align with the field.
Explanation:
Ferromagnetic Materials:
Ferromagnetic materials, like iron, exhibit strong magnetic properties due to the alignment of their atomic magnetic moments within small regions called domains.
External Field Interaction:
When an external magnetic field is applied, it exerts a force on these magnetic domains, causing them to reorient and align with the applied field.
Domain Growth and Shrinkage:
Domains aligned with the external field tend to grow in size, effectively increasing the overall magnetization of the material. Conversely, domains aligned against the field tend to shrink.
A long, straight wire carries a current \(i\). The magnetizing field intensity \(H\) is measured at a point \(P\) close to the wire. A long, cylindrical iron rod is brought close to the wire so that the point \(P\) is at the centre of the rod. The value of \(H\) at \(P\) will
(c) From the Biot-Savart law, magnetic field (\(B\)) at a point \(P\) close to the wire carrying current i is given by, \(\vec{B}=\frac{\mu_0}{4 \pi} \frac{i \overrightarrow{d l} \times \vec{r}}{r^3}\)
Magnetising field intensity (\(H\)) will be,
\(
H=\frac{B}{\mu_0}=\frac{1}{4 \pi} \frac{i \overrightarrow{d l} \times \vec{r}}{r^3}
\)
Now, as the cylindrical rod is brought close the wire such that centre of the rod is at \(P\) , then distance of point \(P\) from the wire(r) will remain same. Hence, magnetic field intensity will remain almost constant. Also even when the rod is carrying any current then \(B\) will be zero at the centre of the rod so the value of Magnetising field intensity will remain the same at point \(P\).
The magnetic susceptibility is negative for
(b) Diamagnetic materials only.
Explanation: Diamagnetic materials have a negative magnetic susceptibility, meaning they are weakly repelled by a magnetic field. Paramagnetic and ferromagnetic materials have positive susceptibilities and are attracted to magnetic fields.
Why other options are incorrect:
(a) paramagnetic materials only: Paramagnetic materials have a positive magnetic susceptibility.
(c) ferromagnetic materials only: Ferromagnetic materials have a positive magnetic susceptibility.
(d) paramagnetic and ferromagnetic materials: Both paramagnetic and ferromagnetic materials have positive susceptibilities.
The desirable properties for making permanent magnets are
(a) High retentivity and high coercive force.
Explanation:
High retentivity:
This means the magnet can retain a strong magnetic field even after the external magnetizing field is removed. A high retentivity ensures the magnet stays magnetized for a long time.
High coercivity:
This refers to the strength of the external magnetic field needed to demagnetize the magnet. A high coercivity means the magnet is resistant to demagnetization by external fields.
Electromagnets are made of soft iron because soft iron has
(d) Low retentivity and low coercive force.
Explanation:
Retentivity:
This refers to the ability of a material to retain its magnetization after the magnetizing field is removed. A low retentivity means the magnet quickly loses its magnetism when the current is stopped, which is desirable for electromagnets.
Coercive force:
This is the strength of the magnetic field required to demagnetize a material. A low coercive force means the magnet can be easily demagnetized with a relatively small field, again, a desirable property for electromagnets.
Pick the correct options.
(a, b) The correct options are (a) All electrons have magnetic moment and (b) All protons have magnetic moment.
Explanation:
Electrons:
Electrons have a magnetic moment due to their spin, which is an inherent property. They also have an orbital magnetic moment based on their movement around the nucleus.
Protons:
Protons, being positively charged particles, also possess a magnetic moment.
The permanent magnetic moment of the atoms of a material is not zero. The material
(d) The correct answer is may be paramagnetic.
Explanation:
Paramagnetic materials:
These materials have individual atoms with permanent magnetic moments, but they are randomly oriented in the absence of an external magnetic field. When an external field is applied, these moments align partially with the field, causing a slight net magnetization.
Diamagnetic materials:
In diamagnetic materials, all electrons are paired, resulting in no net magnetic moment per atom. When exposed to an external magnetic field, they develop a very weak magnetic moment that opposes the applied field, causing them to be slightly repelled.
Ferromagnetic materials:
These materials also have permanent magnetic moments, but unlike paramagnetic materials, their moments are aligned even without an external field. This strong alignment allows them to retain their magnetization after the external field is removed.
The permanent magnetic moment of the atoms of a material is zero. The material
(b) Must be diamagnetic.
Explanation:
A diamagnetic material has no permanent magnetic moment because all its electrons are paired, resulting in a zero net magnetic moment per atom.
Why other options are incorrect:
(a) must be paramagnetic:
Paramagnetic materials have a small, non-zero permanent magnetic moment due to unpaired electrons.
(c) must be ferromagnetic:
Ferromagnetic materials have a large, permanent magnetic moment because their electrons are aligned in parallel, even without an external magnetic field.
(d) may be paramagnetic:
This option is incorrect because paramagnetic materials have a permanent magnetic moment, while the question specifies that the material has a zero permanent magnetic moment.
Which of the following pairs has quantities of the same dimensions?
(c, d) Dimension of Magnetic field B is given by,
\(
\begin{aligned}
& \mathrm{F}=\mathrm{Bqv} \\
& \Rightarrow B=\frac{F}{q v} \\
& \Rightarrow \frac{\left[M L T^{-2}\right]}{[A T]\left[L T^{-1}\right]}=\left[M T^{-2} A^{-1}\right]
\end{aligned}
\)
Dimension of magnetising field intensity H is given by,
\(
\begin{aligned}
& \mathrm{H}=\left[\frac{I d l}{r^2}\right] \\
& \Rightarrow \frac{[A L]}{\left[L^2\right]}=\left[A L^{-1}\right]
\end{aligned}
\)
Dimension of intensity of magnetisation I is also \(\left[A L^{-1}\right]\) as both Magnetising field intensity H and intensity of magnetisation I are measured in \(\mathrm{Am}^{-1}\).
Hence, option (a) and option (b) is incorrect. Option (c) is correct.
As Longitudinal strain and magnetic susceptibility both are dimensionless quantity. Hence, option (d) is correct.
Mark out the correct options.
(a, d) When a material is placed in magnetic field, dipole moment are induced in the atoms by the applied magnetic field. Since the direction of magnetic field due to induced dipole moment is opposite to the applied magnetic field. Therefore, resultant magnetic field is smaller than the applied magnetic field. This process is called diamagnetism. As this process takes place for all the material, therefore all the material exhibit diamagnetism. Hence, option (a) and (d) are correct.
Diamagnetic material do not have permanent magnetic moment on their own. When they are placed in magnetic field, dipole moments are induced by the applied magnetic field. Thus, there is no net alignment of permanent magnetic moment so these mterials do not have any permanenet magnetic momentof their own. Hence, option (b) is incorrect.
Magnetic field intensity is not zero in free space. Hence, option (c) is incorrect.
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