A rod of length l having uniformly distribute | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation. The charge on this element is $dq = \frac{Q}{l} dx$.Since the

Vedclass · Accepted Answer

$\frac{\pi f Q l^2}{3}$

Vedclass · Answer

(B) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation. The charge on this element is $dq = \frac{Q}{l} dx$.Since the rod rotates with frequency $f$,the current $di$ associated with this rotating charge element is $di = dq \cdot f = \frac{Q}{l} f dx$.The magnetic moment $d\mu$ of this circular loop of radius $x$ is given by $d\mu = (di) \cdot A$,where $A = \pi x^2$ is the area of the loop.Substituting the values,we get $d\mu = \left( \frac{Q f}{l} dx \right) (\pi x^2) = \frac{\pi f Q}{l} x^2 dx$.To find the total magnetic moment $\mu$,we integrate $d\mu$ from $x = 0$ to $x = l$:$\mu = \int_0^l \frac{\pi f Q}{l} x^2 dx = \frac{\pi f Q}{l} \left[ \frac{x^3}{3} \right]_0^l = \frac{\pi f Q}{l} \cdot \frac{l^3}{3} = \frac{1}{3} \pi f Q l^2$.

$A$ rod of length $l$ having uniformly distributed charge $Q$ is rotated about one end with constant frequency $f$. Its magnetic moment is:

Similar Questions

$A$ charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v.$ The associated magnetic moment $\mu$ is given by

$A$ current carrying wire of length $L$ is transformed into a coil of $N$ turns. To achieve maximum magnetic moment,the coil

Show the magnetic field lines for a current-carrying loop. Write the rule for finding the direction of a magnetic field due to a circular current loop.

An electron of charge $e$ moves in a circular orbit of radius $r$ around the nucleus at a frequency $v$. The magnetic moment associated with the orbital motion of the electron is

$A$ circular coil of radius $4\, cm$ and of $20$ turns carries a current of $3$ amperes. It is placed in a magnetic field of intensity $0.5\, Wb/m^2$. The magnetic dipole moment of the coil is.......$A \cdot m^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module