Solveeit Logo

Question

Question: A circular loop of radius \(R\) carries current \({I_2}\) in a clockwise direction as shown in figur...

A circular loop of radius RR carries current I2{I_2} in a clockwise direction as shown in figure. The centre of the loop is a distance DD above a long straight wire. What are the magnitude and direction of the current I1{I_1} in the wire if the magnetic field at the centre of loop is zero?

Explanation

Solution

We can use the Biot-Savart law to find the magnetic field due to a current-carrying loop which is the place at some distance from another current-carrying wire.

Complete step by step solution:
Let us consider that a circular loop of radius RR
current in the loop is I2{I_2}
direction of current is clockwiseclockwise
the distance between center of loop to the straight wire is DD,
and magnetic field at the center of the loop is zerozero
At the centre of the circular loop the current I2{I_2} generates a magnetic field that give by cross \otimes . So the current I1{I_1} must point to the right.
Magnetic field is given by
BA=μoI22R{B_A} = \dfrac{{{\mu _o}{I_2}}}{{2R}} …………...( 1)
And
BB=μoI12πD{B_B} = \dfrac{{{\mu _o}{I_1}}}{{2\pi D}} ………...( 2)
Comparing both magnetic field equations, we get
μoI22R=μoI12πD\dfrac{{{\mu _o}{I_2}}}{{2R}} = \dfrac{{{\mu _o}{I_1}}}{{2\pi D}}
For complete cancellation the two fields’ magnitude, we get
I1=πDI2R{I_1} = \dfrac{{\pi D{I_2}}}{R}

**Hence, The magnitude of the current I1{I_1} in the wire is I1=πDI2R{I_1} = \dfrac{{\pi D{I_2}}}{R} and the direction of the current in the straight wire is left to right when current flowing in the wire is clockwise and magnetic field at center is zero. **

Note: When the current is straight which means the current is passing through a straight wire. The magnetic field produced due to current through a straight conductor is in the form of a concentric circle.