In the figure shown,a current I_1 is establis | vedclass

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

Question

(B) The magnetic field $B$ produced by the long straight wire $AB$ at any point in the plane of the paper is directed perpendicularly inwards (into th

Vedclass · Accepted Answer

towards left

Vedclass · Answer

(B) The magnetic field $B$ produced by the long straight wire $AB$ at any point in the plane of the paper is directed perpendicularly inwards (into the paper) according to the Right-Hand Thumb Rule.For any small current element $Idl$ on the wire $CD$,the force is given by $dF = I(dl 	imes B)$.Since the magnetic field $B$ is uniform in direction (perpendicular to the plane) and the current $I_2$ flows through the wire $CD$,we can consider the net force on the wire $CD$ as the vector sum of forces on its segments.Due to the symmetry of the wire $CD$ with respect to the magnetic field produced by $AB$,the horizontal components of the forces on the two segments of $CD$ cancel each other out.The vertical components of the forces on both segments of $CD$ are directed towards the wire $AB$ (leftwards) because the current in both segments of $CD$ has a component parallel to $AB$ or is oriented such that the cross product $dl 	imes B$ results in a force towards the left.Therefore,the net force on the wire $CD$ is directed towards the left.

In the figure shown,a current $I_1$ is established in the long straight wire $AB$. Another wire $CD$ carrying current $I_2$ is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire $AB$. The force on the wire $CD$ is:

Similar Questions

$A$ straight wire of mass $200 \;g$ and length $1.5 \;m$ carries a current of $2 \;A$. It is suspended in mid-air by a uniform horizontal magnetic field $B$ (Figure). What is the magnitude of the magnetic field (in $T$)?

Find the force per unit length at point $P$. The current in the wire is $5 \, A$ and the distance between the parallel segments is $10 \, cm$.

If the distance between two current-carrying parallel wires is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the magnitude of the force between them (the same current is passing through both wires) is:

For the circuit shown in the figure,the direction and magnitude of the force on the segment $PQR$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module