A bar magnet with a magnetic moment 50 $Am ^2$ is placed in parallel position relative to a magnetic field of 0.4T. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is
\(u=−MBcosθ\)
\(W=Δu\)
\(W=−MBcos180^0(−mBcos0^0)\)
\(W=2MB\)
\(W=2×5×0.4\)
\(W=4J\)
Therefore, the correct answer is (A): 4J
The potential energy of a magnetic moment in a magnetic field is given by:
\[ u = -MB \cos \theta \]
where \( M \) is the magnetic moment, \( B \) is the magnetic field, and \( \theta \) is the angle between the magnetic moment and the magnetic field.
The work done in changing the position is:
\[ W = \Delta u = -MB \cos 180^\circ - (-MB \cos 0^\circ) = 2MB \times (0.4) = 4 \, \text{J} \]
List I (Material) | List II (Subsceptibility(χ)) | ||
A. | Diamagnetic | I. | χ = 0 |
B. | Ferromagnetic | II. | 0 > χ ≥ –1 |
C. | Paramagnetic | III. | χ >> 1 |
D. | Non-magnetic | IV. | 0 < χ < ε (a small positive number) |
Moving charges generate an electric field and the rate of flow of charge is known as current. This is the basic concept in Electrostatics. Another important concept related to moving electric charges is the magnetic effect of current. Magnetism is caused by the current.
Region in space around a magnet where the Magnet has its Magnetic effect is called the Magnetic field of the Magnet. Let us suppose that there is a point charge q (moving with a velocity v and, located at r at a given time t) in presence of both the electric field E (r) and the magnetic field B (r). The force on an electric charge q due to both of them can be written as,
F = q [ E (r) + v × B (r)] ≡ EElectric +Fmagnetic
This force was based on the extensive experiments of Ampere and others. It is called the Lorentz force.