To produce a uniform magnetic field directed | vedclass

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

Question

(B) For a long cylinder with uniform current density $J$,the magnetic field at a point $\vec{r}$ from the axis is given by $\vec{B} = \frac{\mu_0}{2}

Vedclass · Accepted Answer

$\frac{\mu_0 J a}{2}$ along $y$-axis

Vedclass · Answer

(B) For a long cylinder with uniform current density $J$,the magnetic field at a point $\vec{r}$ from the axis is given by $\vec{B} = \frac{\mu_0}{2} (\vec{J} 	imes \vec{r})$.Let the two cylinders have current densities $\vec{J}_1 = J \hat{z}$ and $\vec{J}_2 = -J \hat{z}$.Let $\vec{r}_1$ and $\vec{r}_2$ be the position vectors of a point inside the overlap region relative to the axes of the two cylinders respectively.The magnetic field due to the first cylinder is $\vec{B}_1 = \frac{\mu_0}{2} (\vec{J}_1 	imes \vec{r}_1) = \frac{\mu_0 J}{2} (\hat{z} 	imes \vec{r}_1)$.The magnetic field due to the second cylinder is $\vec{B}_2 = \frac{\mu_0}{2} (\vec{J}_2 	imes \vec{r}_2) = \frac{\mu_0 J}{2} (-\hat{z} 	imes \vec{r}_2) = -\frac{\mu_0 J}{2} (\hat{z} 	imes \vec{r}_2)$.The net magnetic field is $\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 J}{2} [\hat{z} 	imes (\vec{r}_1 - \vec{r}_2)]$.From the geometry,$\vec{r}_1 - \vec{r}_2 = \vec{a}$,where $\vec{a}$ is the vector from the axis of the second cylinder to the first. Given the orientation in the figure,$\vec{a}$ is along the $x$-axis,so $\vec{a} = a \hat{x}$.Thus,$\vec{B}_{net} = \frac{\mu_0 J}{2} (\hat{z} 	imes a \hat{x}) = \frac{\mu_0 J a}{2} \hat{y}$.Therefore,the magnetic field is $\frac{\mu_0 J a}{2}$ along the $y$-axis.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

Similar Questions

An element $dl = dx \hat{i}$ (where,$dx = 1 \, cm$) is placed at the origin and carries a large current $i = 10 \, A$. What is the magnetic field on the $Y$-axis at a distance of $0.5 \, m$?

$A$ $10 \ A$ current is passing through a very long wire of radius $5 \ cm$. The magnetic field at a distance of $2 \ cm$ inside from its curved surface is . . . . . . $\times 10^{-5} \ T$.

If a positive ion is moving away from an observer with some acceleration,then the lines of force of magnetic induction will be

Magnetic induction is a

Vedclass Test Series

Exam Paper Generator

Online Exam Module