In Fg. 5.4 (b). The magnetic needle has magnetic moment `6.7xx10^(-2)Am^(2)` and moment of inertia `=7.5 xx 10^(-6)kg m^(2)`. It performs 10 complete oscillations is 6.70 s. what is the magnitude of the magnetic field?

A magnetic needle has a magnetic moment 5.8xx10^(-2)" A "m^(2) and moment of inertia of 7.8xx10^(-6)"kg m"^(2) , it performs 12 complete oscillations in 6.0 s. What is the magnitude of magnetic field ?

A magnetic needle has magnetic moment of 6*7xx10^-2Am^2 and moment of inertial of 7*5xx10^-6kgm^2 . It perform 10 complete oscillations in 6*70s . What is the magnitude of the magnetic field?

A magnetic needle of magnetic moment 6*7xx10^-2Am^2 and moment of inertia 7*5xx10^-6kgm^2 is performing simple harmonic oscillations in a magnetic field of 0*01T . Time taken for 10 complete oscillation is

A magnetic needle is free to oscillate in a uniform magnetic field as shown in the figure. The magnetic moment of magnetic needle 7.2Am^(2) and moment of inertia I=6.5xx10^(-6)kgm^(2) . The number of oscillation performed in 5s are 10. Calculate the magnitude of magnetic field?

A magnetic dipole of magnetic moment 6xx 10^(-2) A m^2 and moment of inertia 12 xx 10^(-6) kg m^2 performs oscillations in a magnetic field of 2 xx 10^(-2) T. The time taken by the dipole to complete 20 oscillations is (pi~~3)

A bar magnet made of steel has a magnetic moment of 2.5 A m^2 and amass of 6.6 xx 10^(-5) kg . If the density of steel is 7.9 xx 10^3 kg m^(-3) , find the intensity of magnetization of the magnet.

The horizontal component of the earth's magnetic field is 3.6 xx 10^(-5) T where the dip is 60^@ . Find the magnitude of the earth's magnetic field.

A magnet of magnetic moment 10 Am^(2) has magnetic length 5 cm.The strength of magnet is

The magnetic moment of a magnet is 3.6xx10^(-3)A.m^(2) . Its pole strength is 120 mili amp.M . Its magnetic length is

A compass needle made of pure iron (with density 8000 kg//m^3 ) has a length 5 cm, width 1.0 mm and thickness 0.50 mm. The magnitude of magnetic dipole moment of an iron is mu_"Fe"=2 xx 10^(-23) J/T. If magnetisation of needle is equivalent to the alignment of 10% of the atoms in the needle, what is the magnitude of the neddle's magnetic dipole moment mu ? (mass of iron per mole =0.05 kg//"mole", N_A=6xx10^23 )