Magnetic Moment Calculation for Bent Bar
The magnetic moment ($M$) of a bar magnet is typically defined as the product of its pole strength ($q_m$) and its length ($L$), represented vectorially as $\vec{M} = q_m \vec{L}$. We assume the magnetic moment acts along the length of the bar.
Calculating New Magnetic Moment
The original bar has magnetic moment $M$. Let its length be $L$. So, $M = q_m L$.
When the bar is bent into an arc of a circle subtending an angle $\theta = 60^{\circ}$ at the center, the arc length is equal to the original length $L$. The angle in radians is $\theta = 60^{\circ} = \frac{60}{180} \pi = \frac{\pi}{3}$ radians.
Let $r$ be the radius of the circular arc. The arc length is given by $L = r \theta$.
Therefore, $L = r \left(\frac{\pi}{3}\right)$.
This gives the radius as $r = \frac{3L}{\pi}$.
The new magnetic moment, $M'$, is associated with the straight-line distance (chord length, $C$) between the two ends of the bent arc. The magnetic moment vector is considered along this chord.
So, $M' = q_m C$.
The chord length $C$ is calculated using the formula $C = 2r \sin(\frac{\theta}{2})$.
Substituting the values:
$C = 2r \sin\left(\frac{60^{\circ}}{2}\right) = 2r \sin(30^{\circ})$
Since $\sin(30^{\circ}) = \frac{1}{2}$,
$C = 2r \left(\frac{1}{2}\right) = r$.
Now substitute the expression for $r$ back into the equation for $M'$:
$M' = q_m C = q_m r = q_m \left(\frac{3L}{\pi}\right)$
$M' = \frac{3}{\pi} (q_m L)$
Since the original magnetic moment $M = q_m L$, the new magnetic moment is:
$M' = \frac{3M}{\pi}$
Similar Questions
- Consider a fuse wire of length $l$ and radius $r$. The time of heating ($t$) for passing the maximum current will depend on
- A circular coil, carrying current, has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{27}$th of its value at the centre of the coil is
- A resistor of resistance '$R$' draws power '$P$' when connected to an AC source. If an inductance is now placed in series with $R$, such that the impedance of the circuit becomes '$Z$', the power drawn will be
- A simple pendulum of length $l$ has a bob of mass $m$, with a charge $q$. On it a vertical sheet of charge, with surface charge density '$\sigma$' passes through the point of suspension. At equilibrium, if the string makes an angle $\theta$ with the vertical, then
A uniform time-varying magnetic field exists in a circular region of radius $R$, directed perpendicular into the plane of the paper, increasing at a constant rate $\alpha$. A straight conducting rod of length $2R$ is placed exactly along the diameter of the circular region (passing through the centre). Find the induced emf across the rod.
There is a ring of radius $r$ having linear charge density $\lambda$ and rotating with a uniform angular velocity $\omega$. The magnitude of the magnetic field produced by this ring at its own centre would be
($\mu_0 = \text{permeability of air}$)- The displacement current flows through a capacitor when the voltage across its plates
Important Questions from Electricity and Magnetism
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
A wire of length $25 \ m$ and cross-sectional area $5 \ mm^2$ having resistivity of $2 \times 10^{-6} \ \Omega \ m$ is bent into a complete circle. The resistance between diametrically opposite points will be
(DROPPED)
A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is :
First Configuration
The electrostatic potential on the surface of uniformly charged spherical shell of radius $R = 10 \ cm$ is $120 \ V$. The potential at the centre of shell, at a distance $r = 5 \ cm$ from centre, and at a distance $r = 15 \ cm$ from the centre of the shell respectively, are:
The radiation pressure exerted by a $450 \ W$ light source on a perfectly reflecting surface placed at $2m$ away from it, is
