Rankers Physics

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

A cube made of wires of equal length is connected to a battery as shown in the figure. The magnetic field at the centre of the cube is : Magnetic Field Due to Straight Current Carrying Wire diagram: A cube made of wires of equal length
\[\frac{12\mu_{0}I}{\sqrt{2}\pi L}\]
\[\frac{6\mu_{0}I}{\sqrt{2}\pi L}\]
\[\frac{6\mu_{0}I}{\pi L}\]
zero

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A semi-infinite straight conductor carries a current I P is a point at perpendicular distance a from the conductor as shown. The field at P due to the conductor is :
2. 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 :
3. The magnetic field at center O of the arc in figure is :
4. Given below are two statements Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element ($I d \vec{l}$) of the current carrying conductor only. Statement II : Biot-Savart's law is analogous to Coulomb's inverse square law of charge $q$, with the former being related to the field produced by a scalar source, $Id \vec{l}$, while the latter being produced by a vector source, $q$. In light of above statements choose the most appropriate answer from the options given below
5. Assertion (A): The magnetic field induction due to an infinite long current carrying solid cylindrical conductor of radius \(R\), at a distance \(R/2\) and \(2R\) from its axis is same. Reason (R): An infinite long current carrying solid cylindrical conductor is a source of uniform magnetic field.