Solveeit Logo

Question

Question: Same current \(i\) is flowing in three infinitely long wires along positive \(x,y\,and\,z\) directio...

Same current ii is flowing in three infinitely long wires along positive x,yandzx,y\,and\,z directions. The magnetic field at a point(0,0,a)\left( {0,0, - a} \right) would be
A. μ0i2πa(ji)\dfrac{{{\mu _0}i}}{{2\pi a}}\left( {\mathop j\limits^ \wedge - \mathop i\limits^ \wedge } \right)
B. μ0i2πa(ij)\dfrac{{{\mu _0}i}}{{2\pi a}}\left( {\mathop i\limits^ \wedge - \mathop j\limits^ \wedge } \right)
C. μ0i2πa(ij)\dfrac{{{\mu _0}i}}{{2\pi a}}\left( {\mathop i\limits^ \wedge - \mathop j\limits^ \wedge } \right)
D. μ0i2πa(i+j+k)\dfrac{{{\mu _0}i}}{{2\pi a}}\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)

Explanation

Solution

Hint - To solve this question, firstly we should learn about current and magnetic fields and the concepts behind them. And then by using the knowledge we have and the formulas required, we will be easily approaching our answer.

Step-By-Step answer:
Electric current: It is the rate of the flow of electric charge from a point. Electric charge is carried out by the charged particles, therefore electric current is the flow of the charged particles.
Magnetic Field: It is a vector quantity that describes the magnetic influence on an electric charge of other moving charges or magnetized materials. A charge that is moving in a magnetic field experiences a perpendicular force to its own velocity and to the magnetic field.
Let the magnetic field at a point for infinite wire be BB .
Then, here in this case contribution will come from wire along xandyx\,and\,y axis only. Which are perpendicular to each other, apply ampere circuital law and you can find direction and magnitude of BB.

Let both be flowing along in a positive direction. Then, magnetic field at (a)\left( { - a} \right) for xx would be
Bx=μ0I2πaj\overrightarrow {{B_x}} = \dfrac{{{\mu _0}I}}{{2\pi a}}\mathop j\limits^ \wedge
And for yy would be
By=μ0I2πai\overrightarrow {{B_y}} = \dfrac{{{\mu _0}I}}{{2\pi a}} - \mathop i\limits^ \wedge

The net magnetic field would be
B=μ0I2πa(ji)\overrightarrow B = \dfrac{{{\mu _0}I}}{{2\pi a}}\left( {\mathop j\limits^ \wedge - \mathop i\limits^ \wedge } \right)

Hence, the solution to our question is option A - B=μ0I2πa(ji)\overrightarrow B = \dfrac{{{\mu _0}I}}{{2\pi a}}\left( {\mathop j\limits^ \wedge - \mathop i\limits^ \wedge } \right)

Note - Magnetic fields are manifestations of Electromagnetism, one of the four fundamental forces of the universe. Electromagnetism involves the interaction of charged particles. Magnetic fields are generated by moving electric currents, and a moving magnet creates an electric field.
Man-made magnetic fields are used by man in electrical devices such as motors, generators and transformers, Electric transformers, Electromagnets, Electric motors also used in Magnetic Resonance Imaging.