A straight wire of length 20 cm carrying a c | vedclass

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

Question

(D) Given: Length of wire $l = 20 \ cm = 0.2 \ m$,Current $I = \frac{3}{\pi^2} \ A$.When the wire is bent into a circle,its length becomes the circumf

Vedclass · Accepted Answer

$6 	imes 10^{-6} \ T$

Vedclass · Answer

(D) Given: Length of wire $l = 20 \ cm = 0.2 \ m$,Current $I = \frac{3}{\pi^2} \ A$.When the wire is bent into a circle,its length becomes the circumference of the circle: $l = 2 \pi R$.So,$0.2 = 2 \pi R \Rightarrow R = \frac{0.1}{\pi} \ m = \frac{10^{-1}}{\pi} \ m$.The magnetic field $B$ at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.Substituting the values: $B = \frac{(4 \pi 	imes 10^{-7}) 	imes (\frac{3}{\pi^2})}{2 	imes (\frac{10^{-1}}{\pi})}$.Simplifying: $B = \frac{4 \pi 	imes 10^{-7} 	imes 3}{2 	imes 10^{-1} 	imes \pi} = \frac{12 \pi 	imes 10^{-7}}{2 \pi 	imes 10^{-1}} = 6 	imes 10^{-6} \ T$.

$A$ straight wire of length $20 \ cm$ carrying a current of $\frac{3}{\pi^2} \ A$ is bent in the form of a circle. The magnetic field at the centre of the circle is

Similar Questions

In which of the following processes does entropy decrease?

Let $\bar{a} = 3 \hat{i} + 2 \hat{j} + x \hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\bar{a} \times \bar{b}| = r$ is possible,if

If $f_n(x) = \log \log \log \ldots \log x$ (where $\log$ is repeated $n$ times),then $\int (x f_1(x) f_2(x) \ldots f_n(x))^{-1} dx$ is equal to

Let $O$ be the vertex and $Q$ be any point on the parabola $x^2 = 8y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1:3$,then the locus of $P$ is:

The chemical formula of 'tear gas' is

Vedclass Test Series

Exam Paper Generator

Online Exam Module