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Multiple choice 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

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 :

  1. $10.39\times 10^{-5}\ Nm$
  2. $\sqrt{3} \times 10^{-5}Nm$
  3. $6\times 10^{-5} Nm$
  4. $\sqrt{5}\times 10^{-5}Nm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $
$=10^{-6}\times \dfrac{\sqrt{3}}{2}$
$=1.039\times 10^{-4}Nm$
$=10.39\times 10^{-5}Nm$

Multiple choice 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

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

  1. $50\sqrt {2}$
  2. $100$
  3. $100\sqrt {2}$
  4. $150\sqrt {2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The potential energy of a magnetic dipole in a field is U = -M dot B. The change in potential energy is converted into rotational kinetic energy (1/2)I(omega^2). Here, M dot B = (3)(4) + (-4)(3) = 0. However, the system starts from rest and the magnetic torque will cause rotation. The maximum angular velocity occurs when the potential energy is minimized (M aligned with B). The change in potential energy is delta U = M*B - M dot B = |M||B| - 0 = 5*5 = 25 J. Setting (1/2)I(omega^2) = 25 with I = 0.01 gives omega = sqrt(50/0.01) = sqrt(5000) = 50*sqrt(2).

Multiple choice 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

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

  1. 2 d
  2. $ \sqrt {2} $ d
  3. $ d / \sqrt {2} $

  4. d / 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The force between magnetic poles is F = k(m1*m2)/d^2. If m1 becomes 2*m1, to keep F constant, the new distance d' must satisfy (2*m1*m2)/(d')^2 = (m1*m2)/d^2. This simplifies to 2/d'^2 = 1/d^2, so d' = d/sqrt(2).

Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

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

  1. $\sqrt 3J$
  2. $\dfrac {3}{2}J$
  3. $2\sqrt 3J$
  4. $3J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


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


$\Rightarrow MB=2\sqrt 3J$


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

Multiple choice 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

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

  1. 1.1

  2. 0.8

  3. 0.001

  4. 0.01
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $\tau = \bar M \times \bar B$
                $= MB  sin  \theta$
                $= I AB$
                $= 10 \times 0.01 \times 0.1 = 0.01 Nm$.

Multiple choice 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

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^o$, and one of the fields has a magnitude of $1.2\times 10^{-2} T$. If the dipole comes to stable equilibrium at an angle of $15^o$ with this field, what is the magnitude of the other field?

  1. $3\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
  2. $\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
  3. $6\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
  4. $2\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here, $\theta=60^{0}, B1=1.2\times 10^{−2}tesla,$
$\theta _1=15^{0}, \theta _{2}=60^{0}−15^{0}=45^{0}.$
In equilibrium, torque due to two fields must balance i.e.
$\tau _{1}=\tau _{2}$
$MB _{1}sin\theta _1=MB2sin\theta _2$

$\implies 1.2\times 10^{-2}\times sin15^{\circ}=B _2sin(60-15)^{\circ}=B _2sin45^{\circ}$
$\implies B _2=4.4\times 10^{-3}T$

Multiple choice 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

A short bar magnet experiences a torque of magnitude $0.64\ J$. When it is placed in a uniform magnetic field of $0.32\ T$, making an angle of $30^{\circ}$ with the direction of the field. The magnetic moment of the magnet is

  1. $2\ Am^{2}$
  2. $4\ Am^{2}$
  3. $6\ Am^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Torque, $\tau = 0.64\ J, B = 0.32\ T, \theta = 30^{\circ}$
Torque, $\tau = MB\sin \theta$
$0.64 = M\times 0.32\sin 30^{\circ}$
$0.64 = M\times 0.32\times \dfrac {1}{2}$
$M = \dfrac {2\times 0.64}{0.32} = 4\ Am^{2}$.

Multiple choice 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

A short bar magnet placed with its axis at $30^o$ with a uniform external magnetic field of $0.35$ T experiences a torque of magnitude equal to $4.5\times 10^{-2}$N m. The magnitude of magnetic moment of the given magnet is?

  1. $26$J $T^{-1}$
  2. $2.6$J $T^{-1}$
  3. $0.26$J $T^{-1}$
  4. $0.026$J $T^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:
The angle made by the magnetic field is, $θ=30^o$
The magnetic field in the region is, $B=0.35\ T$
The torque acting on the bar magnet is $\tau=4.5\times 10^{-2}\ Nm$

The torque acting on the magnet is given by:

$\tau=MB\ sin\ θ$

$ 4.5\times 10^{-2}= M\times 0.35 \times (sin 30^o)$

$⟹M=0.26\ JT^{-1} $
Multiple choice 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

A bar magnet has a magnetic moment of $200$ A $m^2$. The magnet is suspended in a magnetic field of $0.30$N $A^{-1}m^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^o$, will be:

  1. $30$ N m
  2. $30\sqrt{3}$ N m
  3. $60$ N m
  4. $60\sqrt{3}$ N m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$\tau =MB \sin \theta$
Here, $M=200A m^2, B=0.30N A^{-1}m^{-1}$ and $\theta =30^o$
$\therefore \tau =200\times 0.30\times \sin 30^o$
$\tau =30$N m