Magnetic Properties of Matter — Page 2 | JEE Physics

Q11

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is

2s.

The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be

A

2\sqrt 3 \,s

B

{2 \over 3}\,\,s

C

2\,s

D

{2 \over {\sqrt 3 }}\,s

Correct Answer

Option B

Solution

T = 2\pi \sqrt {{1 \over {M \times B}}}

where

I = {1 \over {12}}m{\ell ^2}

When the magnet is cut into three pieces the pole strength will remain the same and

M.{\rm I}.\left( {I'} \right) = {1 \over {12}}\left( {{m \over 3}} \right){\left( {{\ell \over 3}} \right)^2} \times 3 = {I \over 9}

We have, Magnetic moment

(M)

$=$ Pole strength

\left( m \right) \times \ell

$\therefore$ New magnetic moment,

M' = m \times \left( {{\ell \over 3}} \right) \times 3 = m\ell = M

$\therefore$

T' = {T \over {\sqrt 9 }} = {2 \over 3}s.

Q12

The materials suitable for making electromagnets should have

A high retentivity and low coercivity

B low retentivity and low coercivity

C high retentivity and high coercivity

D low retentivity and high coercivity

Correct Answer

Option B

Solution

Materials suitable for making electromagnets should have low retentivity and low coercivity.

Retentivity (or remanence) is the ability of a magnetic material to retain its magnetism after the removal of the magnetizing force.

For an electromagnet, we want this to be low, as we want the magnet to only be magnetic when current is flowing.

Coercivity is the ability of a magnetic material to resist becoming demagnetized.

Again, for an electromagnet, we want this to be low, as we want to easily turn off the magnetism when the current is removed.

So, the correct option is : Option B: Low retentivity and low coercivity.

Q13

Curie temperature is the temperature above which

A a ferromagnetic material becomes paramagnetic

B a paramagnetic material becomes diamagnetic

C a ferromagnetic material becomes diamagnetic

D a paramagnetic material becomes ferromagnetic

Correct Answer

Option A

Solution

The Curie temperature is the temperature above which a ferromagnetic material becomes paramagnetic.

Ferromagnetic materials have a high degree of magnetization in the presence of a magnetic field.

Above the Curie temperature, these materials lose their ferromagnetic behavior and become paramagnetic, meaning they are weakly attracted to a magnetic field and do not retain any magnetization in the absence of an external magnetic field.

So, the correct option is : Option A : a ferromagnetic material becomes paramagnetic.

Q14

A magnetic needle is kept in a non-uniform magnetic field. It experiences :

A neither a force nor a torque

B a torque but not a force

C a force but not a torque

D a force and a torque

Correct Answer

Option D

Solution

A magnetic needle kept in non uniform magnetic field experience a force and torque due to unequal forces acting on poles.

Q15

Needles

{N_1}

,

{N_2}

and

{N_3}

are made of ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will

A attract

{N_1}

and

{N_2}

strongly but repel

{N_3}

B attract

{N_1}

strongly,

{N_2}

weakly and repel

{N_3}

weakly

C attract

{N_1}

strongly, but repel

{N_2}

and

{N_3}

weakly

D attract all three of them

Correct Answer

Option B

Solution

Ferromagnetic substance has magnetic domains whereas para-magnetic substances have magnetic dipoles which get attracted to a magnetic field.

Diamagnetic substances do not have magnetic dipole but in the presence of external magnetic field due to their orbital motion of electrons these substances are repelled.

Q16

Relative permittivity and permeability of a material

{\varepsilon _r}

and

{\mu _r},

respectively. Which of the following values of these quantities are allowed for a diamagnetic material?

A

{\varepsilon _r} = 0.5,\,\,{\mu _r} = 1.5

B

{\varepsilon _r} = 1.5,\,\,{\mu _r} = 0.5

C

{\varepsilon _r} = 0.5,\,\,{\mu _r} = 0.5

D

{\varepsilon _r} = 1.5,\,\,{\mu _r} = 1.5

Correct Answer

Option B

Solution

For a diamagnetic material, the value of

{\mu _r}

is less than one. For any material, the value of

{ \in _r}

is always greater than

1.

Q17

Two short bar magnets of length

1

cm

each have magnetic moments

1.20

A{m^2}

and

1.00

A{m^2}

respectively. They are placed on a horizontal table parallel to each other with their

N

poles pointing towards the South. They have a common magnetic equator and are separated by a distance of

20.0

cm.

The value of the resultant horizontal magnetic induction at the mid-point

O

of the line joining their centres is close to

\left( \, \right.

Horizontal component of earth's magnetic induction is

3.6 \times 10.5Wb/{m^2})

A

3.6 \times 10.5\,\,Wb/{m^2}

B

2.56 \times 10.4\,\,Wb/{m^2}

C

3.50 \times 10.4\,\,Wb/{m^2}

D

5.80 \times 10.4\,Wb/{m^2}

Correct Answer

Option B

Solution

Given:

{M_1} = 1.20A{m^2}\,\,\,

and

\,\,\,{M_2} = 1.00A{m^2}

r = {{20} \over 2}cm = 0.1m

{B_{net}} = {B_1} + {B_2} + {B_H}

{B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}

= {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}

= 2.56 \times {10^{ - 4}}\,\,wb/{m^2}

Q18

At some location on earth, the horizontal component of earth’s magnetic field is 18 × 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45o angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is -

A 3.6

\times

10

-

5 N

B 1.8

\times

10

-

5 N

C 1.3

\times

10

-

5 N

D 6.5

\times

10

-

5 N

Correct Answer

Option D

Solution

Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth.

Hence, the dip ' $\theta$ ' at this place is $45^{\circ}$ (given).

We know that, horizontal and vertical components of earth's magnetic field $\left(B_H\right.$ and $B_V$ ) are related as

\frac{B_V}{B_H}=\tan \theta

Here, $\theta=45^{\circ}$ and $B_H=18 \times 10^{-6} \mathrm{~T}$

\Rightarrow B_V=B_H \tan 45^{\circ}

\Rightarrow B_V=B_H=18 \times 10^{-6} \mathrm{~T} \quad\left(\because \tan 45^{\circ}=1\right)

Now, when the external force $F$ is applied, so as to keep the needle stays in horizontal position is shown below, Taking torque at point $P$ , we get

\begin{array}{ll} m B_V \times 2 l=F l \\\\ \therefore F=2 \times m B_V \end{array}

Substituting the given values, we get

\begin{aligned} & =2 \times 1.8 \times 18 \times 10^{-6} \\\\ & =6.48 \times 10^{-5}=6.5 \times 10^{-5} \mathrm{~N} \end{aligned}

Q19

A magnet of total magnetic moment 10-2

{\widehat i}

A-m2 is placed in a time varying magnetic field, B

{\widehat i}

(cos

\omega t

) where B = 1 Tesla and

\omega

= 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is -

A 0.014 J

B 0.028 J

C 0.01 J

D 0.007 J

Correct Answer

Option A

Solution

To determine the work done in reversing the direction of a magnetic moment in a time-varying magnetic field, we'll follow these steps: Given: Magnetic moment: $\mathbf{m} = 10^{-2} \hat{i}$ A·m² Magnetic field: $\mathbf{B}(t) = B \cos(\omega t) \hat{i}$ , where $B = 1$ T and $\omega = 0.125$ rad/s Time at which reversal occurs: $t = 1$ s Step 1: Calculate the Magnetic Field at $t = 1$ s $B(t) = B \cos(\omega t) = 1 \times \cos(0.125 \times 1) = \cos(0.125 \text{ rad})$ Compute $\cos(0.125 \text{ rad})$ : $\cos(0.125 \text{ rad}) \approx 0.9922$ So, $B(t) \approx 0.9922 \text{ T}$ Step 2: Calculate the Work Done The potential energy $U$ of a magnetic dipole in a magnetic field is: $U = -\mathbf{m} \cdot \mathbf{B} = -mB \cos\theta$ The work done $W$ in reversing the magnetic moment from $\theta = 0^\circ$ to $\theta = 180^\circ$ is: $W = U_{\text{final}} - U_{\text{initial}} = [-mB \cos(180^\circ)] - [-mB \cos(0^\circ)] = mB (\cos 0^\circ - \cos 180^\circ)$ Simplify: $W = mB (1 - (-1)) = 2mB$ Substitute the values: $W = 2 \times (10^{-2} \text{ A·m}^2) \times (0.9922 \text{ T}) \approx 0.01984 \text{ J}$ Step 3: Approximate Using RMS Value Given that the magnetic field is time-varying, we can consider the root mean square (RMS) value of $\cos(\omega t)$ over a complete cycle: $\text{RMS of } \cos(\omega t) = \dfrac{1}{\sqrt{2}}$ Thus, the RMS value of the magnetic field is: $B_{\text{RMS}} = B \times \dfrac{1}{\sqrt{2}} = 1 \times \dfrac{1}{\sqrt{2}} \approx 0.7071 \text{ T}$ Now, calculate the work done using $B_{\text{RMS}}$ : $W_{\text{RMS}} = 2mB_{\text{RMS}} = 2 \times (10^{-2}) \times 0.7071 \approx 0.01414 \text{ J}$

Q20

A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20

\times

10–6 J/ T when a magnetic intensity of 60

\times

103 A/m is applied. Its magnetic susceptibility is

A 3.3

\times

10–4

B 2.3

\times

10–2

C 4.3

\times

10–2

D 3.3

\times

10–2

Correct Answer

Option A

Solution

x =

{1 \over H}

I =

{{Magnetic\,moment} \over {Volume}}

I =

{{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}

= 20 N/m2 x =

{{20} \over {60 \times {{10}^{ + 3}}}}

=

{1 \over 3} \times {10^{ - 3}}

= 0.33 $\times$ 10 $-$ 3 = 3.3 $\times$ 10 $-$ 4