Tag: magnetic field due to bar magnet

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
$B\propto\dfrac{1}{d^3}$

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

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 $\displaystyle \frac{d}{4}?$

  1. 16B

  2. 32B

  3. 64B

  4. 128B
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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 :

  1. (1/r)
  2. (1/r$^2$)
  3. (1/r$^3$)
  4. (1/r$^5$)
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  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. $\displaystyle \frac{8\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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.

  1. $ 10cm $
  2. $ 20cm $
  3. $ 30cm $
  4. $ 40cm $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=2l = 20 cm$

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

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 :

  1. $2 \times 10^{-4}T$
  2. $2 \times 10^{-5}T$
  3. $3 \times 10^{-5}T$
  4. $3 \times 10^{-4}T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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 $\displaystyle \frac{4d}{5}?$

  1. $\displaystyle \frac{125}{4}B$
  2. $\displaystyle \frac{125}{32}B$
  3. $\displaystyle \frac{125}{64}B$
  4. $\displaystyle \frac{125}{16}B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

A short bar magnet with the north pole facing north forms a neutral point at P in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane, the net magnetic induction at $P$ is ( Horizontal component of earth's magnetic field $= B _H$):

  1. zero
  2. $2 B _H$
  3. $\displaystyle \dfrac{\sqrt{5}}{2} B _H$
  4. $\sqrt{5}B _H$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When the north pole of short bar magnet is facing North pole of the earth, at the neutral point P, which is on equatorial line. 
$B _H = \dfrac {\mu _0 M}{4\pi d^3} = B _1$ ............(1)
When the magnet is rotated by $90^o$, the magnetic induction at P which is on axial line,
$B _H = \dfrac {\mu _0 2M}{4\pi d^3} = B _2$ ............(2)
Therefore, net magnetic induction at P is
$B _{net} = \sqrt {(B _1^2 + B _2^2)}$
$B _{net} = \sqrt {(1^2 + 2^2)}B _H = \sqrt 5 B _H$

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

The magnetic field strength at a point at a distance d from the centre on the axial line of a very short bar magnet of Magnetic moment $M$ is $B$. The Magnetic induction at a distance $2d$ from the centre on the equatorial line of a Magnetic Moment $8M$ will be

  1. $4B$
  2. $0.25B$
  3. $0.5B$
  4. $2B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$B=\frac{2\mu _0m}{4 \pi d^{3}}$

$B equatorial =\frac{\mu _0m}{4 \pi d^{3}}$

$d _1=d$
$d _2=2d$
Also,
$m _1=M$
$m _2=8M$

$B _{1}=\frac{2\mu _0m}{4 \pi d^{3}}............1$

$B _{2}=\frac{2\mu _0m}{4 \pi d^{3}} = \frac{\mu _0 8M}{4 \pi (2d)^{3}}.........2$
From 1 & 2 we get that,
$B _{equatorial} =\frac{B}{2}$

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

Charge is uniformly distributed in  a space. The net flux passing through the surface of an imaginary cube of side''a'' in the spaceis $\phi $ the space is 0. The net flux passing through the surface of an imaginary sphere of radius ''a''- in the space will be:

  1. $\phi $
  2. $\dfrac { 3 }{ 4\pi } \phi $
  3. $\dfrac {2\pi }{ 3 } \phi $
  4. $\dfrac {4\pi }{ 3 } \phi $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

external flux of a surface is given by : E.ds.

since, the flux through the cube would be $E\times a2 = x$

therefore for a sphere,  the flux would be $E.\times Φ a2$

which is equal to $Φ$