A circular coil of radius 10 text{ cm} and 10 | vedclass

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

Question

(C) Given: Radius of the coil $r = 10 	ext{ cm} = 0.1 	ext{ m}$.Number of turns $N = 100$.Current $I = 1 	ext{ A}$.The magnetic moment $M$ of a cur

Vedclass · Accepted Answer

$3.142 	ext{ A m}^{2}$

Vedclass · Answer

(C) Given: Radius of the coil $r = 10 	ext{ cm} = 0.1 	ext{ m}$.Number of turns $N = 100$.Current $I = 1 	ext{ A}$.The magnetic moment $M$ of a current-carrying coil is given by the formula $M = N I A$,where $A$ is the area of the coil.The area $A = \pi r^{2} = \pi 	imes (0.1 	ext{ m})^{2} = 0.01 \pi 	ext{ m}^{2}$.Substituting the values into the formula:$M = 100 	imes 1 	imes (0.01 	imes 3.142) 	ext{ A m}^{2}$.$M = 100 	imes 0.03142 	ext{ A m}^{2} = 3.142 	ext{ A m}^{2}$.Thus,the magnetic moment of the coil is $3.142 	ext{ A m}^{2}$.

$A$ circular coil of radius $10 \text{ cm}$ and $100$ turns carries a current of $1 \text{ A}$. What is the magnetic moment of the coil?

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

$A$ square loop of side $l$ is placed in the neighbourhood of an infinitely long straight wire carrying a current $I_1$. The loop carries a current $I_2$ as shown in the figure.

$A$ wire of length $314 \, cm$ carrying a current of $14 \, A$ is bent to form a circle. The magnetic moment of the coil is $........ \, A \cdot m^2$. [Given $\pi = 3.14$]

Magnetic field at the centre of a circular loop of area $A$ is $B$. The magnetic moment of the loop will be ($\mu_{0} =$ permeability of free space).

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module