Related InterviewSolutions

• A magnetic wire of dipole moment `4pi Am^(2)` is bent in the form of semicircle. The new magnetic moment isA. `4 pi Am^(2)`B. `8pi Am^(2)`C. `4 Am^(2)`D. None of these
• A small magnet of dipole moment `M` is kept on the arm of a deflection magentometer set in tan `A` position at a distance of `0.2 m`. If the deflection is `60^(@)`, find the value of `P (B_(H)=0.4xx10^(-4)T)`.A. `2.77 Am^(2)`B. `8 Am^(2)`C. `0.2 Am^(2)`D. None of these
• A current I flows in a conducting wire of lenth L. If we bent it in a circular form, then calculate its magnetic dipole moment.A. `(ll)/(4pi)`B. `(l^(2)l)/(4pi)`C. `(ll^(2))/(4pi)`D. `(l^(2)l^(2))/(4pi)`
• Two similar equal magnetic poles when separated by a distance of `1 m`,A. `10 Am`B. 20 AmC. 50 AmD. 100 Am
• Calculate the force acting between two small magnets, placed in end on position `0.1 m` apart from their centres. (Given, magnetic moment of each magnet is `5 Am^(2)`).A. 0.6 NB. 0.8 NC. 0.15 ND. 0.2 N
• Among the following properties describing diamagnetism, identify the property that is wrongly stated.A. Diamagnetic material do not have permanent magnetic momentB. Diamagnetism is explained in terms of electromagnetic inductionC. Diamagnetic materials have a small positive susceptibilityD. The magnetic moment of individual electrons neutralise each other
• The magnetic induction inside a solenoid is `6.5xx10^(-4)T`. When it is filled with iron medium, then the induction becomes `1.4 T`. The relative permeability of iron will beA. 1578B. 2355C. 1836D. 2154
• Inside a long solenoid wounded with 300 turns/m, an iron rod is placed. An iron rod is 0.2 m long, 10 mm in diameter and of permeability `10^(3)`. Calculate the magnetic moment of the rod, if 0.5 A of current is passed through the rod.A. 2.356 Sl unitsB. 1.335 Sl unitsC. 3.664 Sl unitsD. 1.664 Sl units
• Calculate the angle of dip, if a dip needle oscillating in a vertical plane makes 40 oscillations per minute in a magnetic meridian and 30 oscillations per minute in a vertical plane at right angle to the magnetic meridian.A. `theta=sin^(-1)(0.5625)`B. `theta=sin^(-1)(0.325)`C. `theta=sin^(-1)(0.425)`D. `theta=sin^(-1)(0.235)`
• In previous problem, the angle of dip isA. `tan^(-1)(0.75)`B. `tan^(-1)(0.5)`C. `tan^(-1)(0.8)`D. None of these