Solenoid Core in Magnetic Field
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Q. An iron rod of susceptibility 599 is subjected to a magnetizing field of 1200 Am−1. The permeability of the material of the rod is :
Take, μ0=4π×10−7 TmA−1
Take, μ0=4π×10−7 TmA−1
- 2.4π×10−3 TmA−1
- 2.4π×10−5 TmA−1
- 2.4π×10−7 TmA−1
- 2.4π×10−4 TmA−1
Q.
What is the unit of magnetic intensity?
Q. An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 Am−1. The permeability of the material of the rod is :(μ0=4π×10−7 T mA−1)
- 2.4π×10−5 T mA−1
- 2.4π×10−7 T mA−1
- 2.4π×10−4 T mA−1
- 8×10−5 T mA−1
Q. 75.Magnetic needle oscillates in a horizontal plane with the time period t at a place with angle of dip 60 degree when the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian its period will be 1)T/2 2)T 3)2T 4)2T
Q. The value of aluminum's susceptibility is 2.2×10−5. The percentage increase in the magnetic field, if space within a current carrying toroid, is filled with aliminum, is x104. Then, the value of x is
Q. A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is
- 30∘
- 45∘
- 60∘
- 90∘
Q. A magnet of magnetic moment 50^iA−m2 is placed along the x - axis in a magnetic field →B=(0.5^i+3.0^j)T.The torque acting on the magnet is
- 175 ^k N−m
- 150 ^k N−m
- 75 ^k N−m
- 25√37 ^k N−m
Q. A bar magnet when placed at an angle of 30∘ to the direction of magnetic field induction of 5×10−2 T, experiences a moment of couple 25×10−6 N−m. If the length of the magnet is 5 cm its pole strength is
- 5×10−2 A−m
- 2 A−m
- 5 A−m
- 2×10−2 A−m
Q.
A magnet of total magnetic moment is placed in a time varying magnetic field, where and . The work done for reversing the direction of the magnetic moment at , is
Q. A bar magnet of magnetic moment 104 J/T is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of 4×10−5 T to a direction 60∘ from the field will be
- 0.2 J
- 2.0 J
- 4.18 J
- 4.2×102 J
Q. The pole strength of a bar magnet is 48 ampere - metre and the distance between its poles is 25 cm. The moment of the couple by which it can be placed at an angle of 30∘ with the uniform magnetic intensity of flux density 0.15 Newton / ampere – metre will be
- None of the above
- 8 Newton - metre
- 12 Newton - metre
- 0.9 Newton - metre
Q. The work done in rotating a magnet of magnetic moment 2 A−m2 in a magnetic field of 5×10−3 T from the direction along the magnetic field to opposite direction to the magnetic field, is
- 2×10−2 J
- 10−2 J
- Zero
- 10 J
Q. A magnet of length 0.1 m and pole strength 10−4 A.m is kept in a magnetic field of 30 Wb/m2 at an angle 30∘. The couple acting on it is ……… ×10−4 Nm.
- 6
- 7.5
- 1.5
- 3
Q. A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 180∘ to deflect the magnet by 30∘ from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through 270∘ to deflect the magnet 30∘ from magnetic meridian. The ratio of magnetic moments of magnets is
- 1 : 5
- 1 : 8
- 5 : 8
- 8 : 5
Q. A magnet of magnetic moment 2JT−1 is aligned in the direction of magnetic field of 0.1 T. What is the net work done to bring the magnet normal to the magnetic field
- 0.1 J
- 0.2 J
- 2 J
- 1 J
Q. A bar magnet of length 10 cm and pole strength 2 Am makes an angle 60∘ with a uniform magnetic field of induction B=50 T. The couple acting on it is-
- 20√3 Nm
- 10√3 Nm
- √3 Nm
- 5√3 Nm
Q. A current carrying circular loop is placed in a uniform magnetic field B, the potential energy is minimum when the magnetic moment of the loop is
- Parallel to B
- Perpendicular to B
- Inclined at an angle of 45∘ to B
- Anti-parallel to B
Q. The torque required to keep a magnet of length 20 cm at 30∘ to a uniform magnetic field is 2×10−5 Nm. The magnetic force on each pole is-
- 2×10−6 N
- 2×10−5 N
- 2×10−4 N
- 2×10−3 N
Q. An ideal solenoid having 40 turns cm−1 has an aluminum core, and carries a current of 2.0 A. Calculate the magnetic field B at the centre. The susceptibility χ of aluminium =2.3×10−5.
- 3.2π×10−4 T
- 1.6π×10−4 T
- 0.8π×10−4 T
- π×10−4 T
Q. A bar magnet of moment −→M=^i+^j is placed in a magnetic field induction →B=3^i+4^j+4^k. The torque acting on the magnet is-
- ^i+^j+^k
- ^i−^j
- ^i+^k
- 4^i−4^j+^k
Q. A wire of length 314 cm is used to make a toroid of cross- sectional radius 2 cm. Find the magnetic field at the inner core of the toroid having average radius of 50 cm and current flowing through it is 2 A.
[wires are closely wounded]
[wires are closely wounded]
- 2×10−5 T
- 0.5×10−5 T
- 1×10−4 T
- 0.25×10−3 T
Q. The couple acting on a bar magnet of pole strength 4 Am, when placed in a uniform magnetic field of intensity 10√3 Am−1, such that axis of the magnet makes an angle 60∘ with the direction of the field is 80×10−6 Nm. Find the distance between the poles of the magnet is,
- 0.6 μm
- 0.3 μm
- 1.3 μm
- 0.9 μm
Q. The magnetic susceptibility of a medium is –0.08, what will be its relative permeability?
- 1.8
- 1.08
- 0.08
- 0.92
Q.
Why is the permeability important?
Q. A small bar magnet of moment M is placed in a uniform field H. If magnet makes an angle of 30∘ with field, the torque acting on the magnet is
- MH
- MH2
- MH3
- MH4
Q.
The magnetic field inside a long solenoid having 50 turns cm−1 is increased from 2.5 ×10−3T to 2.5 T when an iron core of cross-sectional area 4 cm2 is inserted into it. Find (a) the current in the solenoid, (b) the magnetization I of the core and (c) the Pole strength developed in the core.
Q. A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is
- 30∘
- 45∘
- 60∘
- 90∘
Q. An iron rod of volume 10−4 m3 and relative permeability 1000 is placed inside a long solenoid wound with 5 turns/cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.
- 15 Am2
- 20 Am2
- 25 Am2
- 30 Am2
Q. A current carrying loop whose magnetic moment is at 30∘ with a uniform external magnetic field of 0.16 T experiences a torque of magnitude 0.032 Nm. If the coil is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively
- −0.064 J, −0.064 J
- +0.064 J, −0.064 J
- +0.064 J, +0.064 J
- −0.064 J, +0.064 J
Q. The magnetic moment of a solenoid of 100 turns carry a current of 1 A and having an area of 0.01 m2 is
- 20 Am2
- 1 Am2
- 0.1 Am2
- Data insufficient