Tag: magnetic field due to bar magnet

Questions Related to magnetic field due to bar magnet

When a charged particle moves perpendicular to a uniform magnetic field, its 

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. energy and momentum both change.

  2. energy changes but momentum remains unchanged.

  3. momentum changes but energy remains unchanged.

  4. energy and momentum both do not change.

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Correct Option: C
Explanation:

Since the force exerted by the magnetic field is perpendicular to the direction of the particle the speed of the particle cannot change but its velocity changes .

so C option is the correct answer

A proton with kinetic energy K describes a circle of radius r in a uniform magnetic. An $\alpha$- particle with kinetic energy K moving in the same magnetic field will describe a circle of radius? 

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. $\cfrac {r} {2}$

  2. r

  3. $2r$

  4. $4r$

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Correct Option: C
Explanation:

WE KNOW THAT 


charge on proton = q
charge on alpha particle = 2q

mass of proton =m 
mass of alpha particle =4m

for proton

 $ r =\dfrac{m\times v}{q\times B}$


for alpha particle

$ R = \dfrac{{4m}\times v}{2q\times B}$

$ R = \dfrac{2m\times v}{q\times B}$

$ R = 2r$

A magnetic needle is placed parallel to a magnetic field. The amount of work done in rotating the coil by an angle of 60$^0$ is W units. Then, the torque required to keep the needle in the displaced position 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. W

  2. $\sqrt{3}$W

  3. $\left( \sqrt{3} / 2 \right )$W

  4. W/2

Show answer
Correct Option: B
Explanation:

Torque at an angle $\theta  = k sin \theta $


work done to bring from zero to 60 degree $ =W = \int _0^{60}{ k sin \theta d \theta } = {\left | -k cos \theta \right |} _0^{60} = \dfrac{k}{2} $

$ \Rightarrow k = 2W $

torque at 60 degree required to keep the needle stable  $= k sin 60 = 2 W \dfrac{\sqrt 3 }{2} = \sqrt 3 W $ 

A bar magnet of moment $4Am^{2}$ is placed in a non-uniform magnetic field. If the field strength at poles are 0.2 T and 0.22 T then the maximum couple acting on it 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. 0.04Nm

  2. 0.84Nm

  3. 0.4 Nm

  4. 0.44Nm

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Correct Option: B
Explanation:

Average of the two field strengths at the poles is $\frac{0.2 +0.22}{2}=0.21$
Copuling( torque): $MB=4 \times 0.21 : Nm= 0.84:Nm$

A magnet of length $30\ cm$ with pole strength $10\ A-m$ is freely suspended in a uniform horizontal magnetic field of induction $40 \times 10^{-6} T$ . If the magnet is deflected by $60^{o}$ from its equilibrium position, the restoring couple acting on it 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. $10.39\times 10^{-5}\ Nm$ 

  2. $\sqrt{3} \times 10^{-5}Nm$ 

  3. $6\times 10^{-5} Nm$

  4. $\sqrt{5}\times 10^{-5}Nm$

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Correct Option: A
Explanation:

$l=30\ cm$
$P=10\ Am$
$B=40\times 10^{-6}$
$\theta =60^o$
$m=pl$
$\vec{\tau}=\vec{m}\times \vec{B}$
$=mB\sin\theta $
$=10^{-6}\times \dfrac{\sqrt{3}}{2}$
$=1.039\times 10^{-4}Nm$
$=10.39\times 10^{-5}Nm$

A current carrying ring with it center at origin and moment of inertia $1\times 10^{-2} kg-m^{2}$, about an axis passing through its center and perpendicular to its plane, has magnetic moment $\vec {M} = (3\hat {i} - 4\hat {j})A - m^{2}$, at time $t = 0$, a magnetic field $\vec {B} = (4\hat {i} + 3\hat {j})T$ is switched on. Maximum angular velocity of the ring is rad/sec 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. $50\sqrt {2}$

  2. $100$

  3. $100\sqrt {2}$

  4. $150\sqrt {2}$

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Correct Option: A

Two unlike magnetic poles are distance "d" apart, and mutually attract with a force "F". If one of the pole strength is doubled and to maintain the same force between them, the new separation between the poles must 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. 2 d

  2. $ \sqrt {2} $ d

  3. $ d / \sqrt {2} $

  4. d / 2

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Correct Option: C

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

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. a force and a torque

  2. a force but not a torque

  3. a torque but not a force

  4. neither a force nor a torque

Show answer
Correct Option: A
Explanation:

A magnetic needle is a magnetic dipole.
In a non uniform Magnetic field Force on each of the poles will be different in both magnitude and direction.
Due to difference in Magnitude the dipole experiences a Force,
Due to difference in Direction the dipole experiences a Torque.

A magnetic needle suspended parallel to a magnetic field requires $\sqrt 3 J$ of work to turn it through $60^o$. The torque needed to maintain the needle in this position 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. $\sqrt 3J$

  2. $\dfrac {3}{2}J$

  3. $2\sqrt 3J$

  4. $3J$

Show answer
Correct Option: D
Explanation:

$W = MB(cos\theta _1 cos \theta _2) = MB (cos 0^o cos 60^o)$


$=MB\left (1-\dfrac {1}{2}\right )=\dfrac {MB}{2}$


$\Rightarrow MB=2\sqrt 3J$


$\tau=MB sin 60^o=(2\sqrt 3)\left (\dfrac {\sqrt 3}{2}\right )J=3J$

A current loop of area 0.01 m$^2$ and carrying a current of 10 A is held perpendicular to a magnetic field of intensity 0.1 T. The torque acting on the loop (in 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. 1.1

  2. 0.8

  3. 0.001

  4. 0.01

Show answer
Correct Option: D
Explanation:

$A = 0.01 m^2, I = 10 A, B = 0.1 T, \theta = 90^o, sin \theta = 1$
Now magnetic moment M =$I \times A$
Torque, $\tau = \bar M \times \bar B$
                $= MB  sin  \theta$
                $= I AB$
                $= 10 \times 0.01 \times 0.1 = 0.01 Nm$.