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Questions Related to physics

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

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

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 $= B _H$):

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

  1. zero

  2. $2 B _H$

  3. $\displaystyle \dfrac{\sqrt{5}}{2} B _H$

  4. $\sqrt{5}B _H$

Show answer
Correct Option: D
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
$B _{net} = \sqrt {(B _1^2 + B _2^2)}$
$B _{net} = \sqrt {(1^2 + 2^2)}B _H = \sqrt 5 B _H$

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

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

  1. $4B$

  2. $0.25B$

  3. $0.5B$

  4. $2B$

Show answer
Correct Option: C
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,
$B _{equatorial} =\frac{B}{2}$

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:

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

  1. $\phi $

  2. $\dfrac { 3 }{ 4\pi } \phi $

  3. $\dfrac {2\pi }{ 3 } \phi $

  4. $\dfrac {4\pi }{ 3 } \phi $

Show answer
Correct Option: A
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 $Φ$

A closely wound solenoid of $2000$ turns and area of cross-section $1.5\times10^{-4}\ m^{2}$ carries a current of $2.0A$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5\times10^{-2}$ Tesla making an angle of $30^{o}$ with the axis of the solenoid. The torque on the solenoid will be-

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

  1. $1.5\times10^{-3}\ N.m$

  2. $1.5\times10^{-2}\ N.m$

  3. $3\times10^{-2}\ N.m$

  4. $3\times10^{-3}\ N.m$

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Correct Option: D

Two bar magnet are kept together and suspended freely in earth's magnetic field. When both like poles are aligned, the time period is 6 sec. When opposite poles are aligned, the time period is 12 sec. The ratio of magnetic moments of the two magnets is :

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

  1. $5/3$

  2. $2/1$

  3. $3/2$

  4. $3/1$

Show answer
Correct Option: A
Explanation:

$T _1= 6 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$T _2= 12 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$\therefore \dfrac{6}{12}=\dfrac{1}{2}= \sqrt{\dfrac{M- M _2}{M _1+M _2}}$ or $ M _1+M _2= 4( M _1-M _2) $
$ 3M _1= 5M _2$
$\dfrac{M _1}{M _2}=\dfrac{5}{3}$

Puneet peddles a stationary bicycle. The peddles are attached to a $100$ turn coil of area $0.10 \ m^2$. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of $0.01 \ T$ perpendicular to the axis of rotation of the coil. What is maximum voltage generated in the coil? 

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

  1. $0.314 \ V$

  2. $0.615 \ V$

  3. $0.921 \ V$

  4. $0.084 \ V$

Show answer
Correct Option: A

A bar magnet of length $16 cm$ has a pole strength of 500 milli amp.m. The angle at which it should be placed to the direction of external magnetic field of induction $2.5$ gauss so that it may experience a torque of $\sqrt { 3 } \times{ 10 }^{ -5 }$ N.m. is

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

  1. $\pi $

  2. $\dfrac { \pi }{ 2 } $

  3. $\dfrac { \pi }{ 3 } $

  4. $\dfrac { \pi }{ 6 } $

Show answer
Correct Option: D