Two parallel wires of equal lengths are separ | vedclass

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

Question

(D) The currents are in opposite directions,so their magnetic fields at the mid-point are in the same direction.Both wires are at a distance $r = 1.5

Vedclass · Accepted Answer

$\frac{5 \mu_{0}}{2 \pi}$

Vedclass · Answer

(D) The currents are in opposite directions,so their magnetic fields at the mid-point are in the same direction.Both wires are at a distance $r = 1.5 \,m$ from the mid-point.The magnetic field due to the first wire is $B_{1} = \frac{\mu_{0} I_{1}}{2 \pi r} = \frac{\mu_{0}}{2 \pi} \cdot \frac{3}{1.5} = \frac{2 \mu_{0}}{2 \pi}$.The magnetic field due to the second wire is $B_{2} = \frac{\mu_{0} I_{2}}{2 \pi r} = \frac{\mu_{0}}{2 \pi} \cdot \frac{4.5}{1.5} = \frac{3 \mu_{0}}{2 \pi}$.Since the fields are in the same direction,the resultant magnetic field is $B = B_{1} + B_{2} = \frac{2 \mu_{0}}{2 \pi} + \frac{3 \mu_{0}}{2 \pi} = \frac{5 \mu_{0}}{2 \pi}$.

Similar Questions

Shown in the figure is a conductor carrying a current $I$. The magnetic field intensity at the point $O$ (common centre of all the three arcs) is

$A$ circular coil carrying current $I$ has a radius $r$ and $n$ turns. The magnetic field along the axis of a coil at a distance $x = 2\sqrt{2}r$ from its centre is (where $\mu_0$ is the permeability of free space).

Find the magnetic field at point $P$ in the given diagram. The wire carries a current $i$.

An electron moving in a circular orbit of radius $r$ makes $n$ rotations per second. The magnetic field produced at the centre has magnitude

$A$ current $I$ flows in a wire bent in the shape of a quarter ring of radius $R$ as shown in the figure. The magnetic field induction $B$ at point $O$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module