Rankers Physics

Magnetic Field Due to Straight Current Carrying Wire: Practice Problem & Solution

A long straight, hollow, conductor (tube) carrying a current has two sections A and C of unequal cross sections joined by a conical section B. 1, 2 and 3 are points on a line parallel to the axis of the conductor. The magnetic fields at 1, 2 and 3 have magnitudes B1, B2 and B3. Then : Magnetic Field Due to Straight Current Carrying Wire diagram: A long straight, hollow, conductor (tube) carrying a
B1 = B2 = B3
B1 = B2 ≠ B3
B1 < B2 < B3
B2 cannot be found unless the dimensions of the section B are known

No solution provided for this question.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Tesla is the unit of (1997, 88)
2. A wire carrying a current i is placed in a uniform magnetic field in the form of the curve y =a sin (πx/L), 0 ≤ x ≤ 2L. The force acting on the wire is :
3. Two long parallel wires are at a distance 2d apart. The carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line XX' is given by :
4. A ring of radius r is carrying a current I. The magnetic field B is always perpendicular to the ring as shown in Fig. The force on the ring is :
5. A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic vector field \[\overrightarrow{B}\] at a point having coordinates (x, y) in the z = 0 plane is :