If the strength of the magnetic field produce | vedclass

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

Question

(A) The magnetic field $B$ at a distance $r$ from an infinitely long straight current-carrying conductor is given by the formula:$B = \frac{\mu_0}{4\p

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The magnetic field $B$ at a distance $r$ from an infinitely long straight current-carrying conductor is given by the formula:$B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r}$Given:$B = 10^{-5}\,Wb/m^2$$r = 10\,cm = 0.1\,m = 10^{-1}\,m$$\frac{\mu_0}{4\pi} = 10^{-7}\,T\cdot m/A$Substituting the values into the formula:$10^{-5} = 10^{-7} 	imes \frac{2i}{10^{-1}}$$10^{-5} = 10^{-7} 	imes 2i 	imes 10^1$$10^{-5} = 2i 	imes 10^{-6}$$2i = \frac{10^{-5}}{10^{-6}}$$2i = 10$$i = 5\,A$Therefore,the current flowing in the conductor is $5\,A$.

If the strength of the magnetic field produced $10\,cm$ away from an infinitely long straight conductor is ${10^{ - 5}}\,Wb/m^2$,the value of the current flowing in the conductor will be........$A$.

Similar Questions

What is the magnetic field at point $P$ in the given figure?

An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to

What should be the current $i$ (in $A$) in a circular coil of radius $5\,cm$ to annul the horizontal component of Earth's magnetic field ${B_H} = 5 \times {10^{ - 5}}\,T$?

In the figure,find out the magnetic field at the center $B$ (Given $I = 2.5 \; A, r = 5 \; cm$).

The magnetic field at the centre of a circular coil of radius $R$ due to current $i$ flowing through it is $B$. The magnetic field at a point along the axis at a distance $R$ from the centre is

Vedclass Test Series

Exam Paper Generator

Online Exam Module