Tag: magnetism

Questions Related to magnetism

When the atomic magnetic moments are randomly oriented in a solid its magnetic behavior is termed as

physics magnetism and matter paramagnetic material classification of magnetic material magnetism

  1. semi-magnetic.

  2. anti-ferromagnetic.

  3. paramagnetic.

  4. poly crystalline.

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Correct Option: C
Explanation:

When the atomic magnetic moments are randomly oriented in a solid its magnetic behavior is termed as paramagnetic.in paramagnetism electrons are unpaired.

Liquid oxygen remains suspended between two pole forces of a magnet because it is:

physics magnetism and matter paramagnetic material classification of magnetic material magnetism

  1. diamagnetic

  2. paramagnetic

  3. ferromagnetic

  4. antiferromagnetic

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Correct Option: B
Explanation:

Oxygen is paramagnetic in nature. So if it will be attracted both by North pole or South pole. When it is placed exactly between two magnetic poles, the forces acting on it due to magnetic poles will be equal & opposite. Hence it will remain suspended between them.

A substance is placed in a non uniform magnetic field. It experiences weak force towards the strong field. The substance is _________ type.

physics magnetism and matter paramagnetic material classification of magnetic material magnetism

  1. Ferromagnetic

  2. Diamagnetic

  3. Paramagnetic

  4. None of these

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Correct Option: C
Explanation:

Paramagnetic substance has the property that when they are placed in an external magnetic field, they get weakly magnetised in the same direction as that of external magnetic field and are weakly attracted by the external magnetic field.
So, correct answer is option C.

Magnetic induction due to a short bar magnet on its axial line is inversely proportional to cube of distance of the point.

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. True

  2. False

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Correct Option: A
Explanation:

Magnetic induction due to a short bar magnet on its axial line,

$B=\dfrac{\mu _0 M}{4\pi d^3}$
Magnetic induction due to a short bar magnet on its axial line is inversely proportional to cube of distance of the point.
$B\propto\dfrac{1}{d^3}$

The magnetic induction due to short bar magnet on its axial line at a distance 'd' is 'B'. What is the magnetic induction due to the same bar magnet on the same line at a distance $\displaystyle \frac{d}{4}?$

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. 16B

  2. 32B

  3. 64B

  4. 128B

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Correct Option: C
Explanation:

$B \displaystyle = \frac{\mu _0  2M}{4 \pi  d^3}$
At $\displaystyle \frac{d}{4} $ distance,
$B' \displaystyle = \frac{\mu _0 2M}{4 \pi (d/4)^3}$
$\displaystyle = \frac{\mu _0  2M}{4 \pi d^3} \times 64 = 64 B$

If r be the distance of a point on the axis of a bar magnet from its centre, the magnetic field at this point is proportional to :

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. (1/r)

  2. (1/r$^2$)

  3. (1/r$^3$)

  4. (1/r$^5$)

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Correct Option: C
Explanation:

For a short Bar Magnet, the magnetic induction at a point on the axix at a distance $r$ from centre is given by  the formula

$B = $   $(\dfrac{\mu _0}{4\pi} )\dfrac{2M}{r^3}$

$\Rightarrow$ $B= \dfrac{K}{r^3}$

$\Rightarrow$ $B\propto \dfrac{1}{r^3}$
Therefore, C is correct option.

A bar magnet of magnetic moment 'M' has a magnetic length '2d'. Find magnetic induction on its equatorial line at a distance $'\sqrt{13 d}'$.

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $\displaystyle \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

  2. $\displaystyle \frac{2\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

  3. $\displaystyle \frac{4\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

  4. $\displaystyle \frac{8\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

Show answer
Correct Option: A
Explanation:

$r = \sqrt{13} d ;  2l  = 2d$
$B _eq = \displaystyle \frac{\mu _0}{4 \pi} \times \frac{\mu}{(r^2 + 1^2)^{\frac{3}{2}}}$
$\displaystyle =\frac{\mu _0}{4\pi} \times \frac{M}{\left ((\sqrt{13}d)^2 + (2d)^2 \right )^{\frac{3}{2}}}$
$=\displaystyle \frac{\mu _0}{4\pi} \times \frac{M}{(17 d^2)^{\frac{3}{2}}} = \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

If ratio of magnetic induction on the axial line of a long magnet at distance 20 cm and 30 cm is 128 : 27. Find length of the magnet.

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $ 10cm $

  2. $ 20cm $

  3. $ 30cm $

  4. $ 40cm $

Show answer
Correct Option: B
Explanation:

$B _{axial} = \displaystyle \frac{\mu _0}{4\pi} \frac{2Mr}{(r^2-l^2)^2}$
$B _{20} : B _{30} = 128  :  27$
$\displaystyle \frac{20}{(20^2 - l^2)^2} \times \frac{(30^2 - l^2)^2}{30} = \frac{128}{127}$
$2 (30^2 - l^2)^2 (27) = 3 (20^2 - l^2)^2 128$
$\sqrt{54}(900 - l^2) = \sqrt{384} (400 - l^2)$
$900 \sqrt{54} - \sqrt{54}l^2 = \sqrt{384} \times 400 - \sqrt{384}l^2$
$l^2 (12.2474) = 12224. 75$
$l^2 = 100 ;  l = \sqrt{100}-10 cm$
$\therefore$ Magnetic length $=2l = 20 cm$

A magnetic induction due to a short bar magnet of magnetic moment 5.4 A m$^2$ at a distance of 30 cm on the equatorial line is :

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $2 \times 10^{-4}T$

  2. $2 \times 10^{-5}T$

  3. $3 \times 10^{-5}T$

  4. $3 \times 10^{-4}T$

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Correct Option: B
Explanation:

$B _{equi} = \displaystyle \frac{\mu _0}{4 \pi } \frac{\mu}{r^3} = \frac{10^{-7}\times 5.4}{(0.3)^3} = 2 \times 10^{-5}T$

The magnetic induction due to short bar magnet on its axial line at a distance 'd' is 'B'. What is the magnetic induction due to the same bar magnet on the same line at a distance $\displaystyle \frac{4d}{5}?$

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $\displaystyle \frac{125}{4}B$

  2. $\displaystyle \frac{125}{32}B$

  3. $\displaystyle \frac{125}{64}B$

  4. $\displaystyle \frac{125}{16}B$

Show answer
Correct Option: C
Explanation:

$B = \displaystyle \frac{\mu _0  2M}{4 \pi  d^3}$
At $(4d/5) $ distance
$B' = \displaystyle \frac{\mu _0  2M}{4 \pi \left ( \frac{4d}{5} \right )^3} = \frac{\mu _0  2M}{4  \pi (d^3)} \left ( \frac{125}{64} \right )$
$=\displaystyle \frac{125}{64}B$