In the above figure,the magnetic field at poi | vedclass

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

Question

(A) The magnetic field at point $C$ is the vector sum of the fields produced by the three segments of the wire: two semi-infinite straight wires and o

Vedclass · Accepted Answer

$\frac{{\mu _0}i}{{4\pi r}}\left[ {\left( {1 + \pi } \right)\hat k - \hat i} \right]$

Vedclass · Answer

(A) The magnetic field at point $C$ is the vector sum of the fields produced by the three segments of the wire: two semi-infinite straight wires and one semi-circular arc.$1$. For the semi-infinite straight wire along the $x$-axis (carrying current towards the origin),the field at $C$ (at distance $r$) is given by $B_1 = \frac{\mu_0 i}{4\pi r} (-\hat{i})$.$2$. For the semi-circular arc of radius $r$,the field at the center is $B_2 = \frac{\mu_0 i}{4r} \hat{k}$.$3$. For the semi-infinite straight wire starting from the origin (carrying current away from the origin),the field at $C$ is $B_3 = \frac{\mu_0 i}{4\pi r} \hat{k}$.Summing these: $\vec{B}_C = B_1 + B_2 + B_3 = \frac{\mu_0 i}{4\pi r} (-\hat{i}) + \frac{\mu_0 i}{4r} \hat{k} + \frac{\mu_0 i}{4\pi r} \hat{k}$.Factoring out $\frac{\mu_0 i}{4\pi r}$,we get: $\vec{B}_C = \frac{\mu_0 i}{4\pi r} [ (1 + \pi) \hat{k} - \hat{i} ]$.

In the above figure,the magnetic field at point $C$ will be:

Similar Questions

Two concentric coils each of radius equal to $2\pi \, cm$ are placed at right angles to each other. $3 \, A$ and $4 \, A$ are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^2$ at the centre of the coils will be $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$.

What is the ratio of the magnetic field at the center of a circular ring of radius $R$ to the magnetic field at a point on its axis at a distance of $3R$ from the center?

In an ionised sodium atom,an electron is moving in a circular path of radius $r$ with angular velocity $\omega$. The magnetic induction in $Wb/m^2$ produced at the nucleus will be

$A$ coil having $9$ turns carrying current $I$ produces a magnetic field $B_1$ at the centre. If the same wire is rewound into a coil of $3$ turns carrying the same current $I$,then the magnetic field at the centre $B_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module