The magnetic field due to a current carrying circular loop of radius $3m$ at a point on the axis at a distance of $4m$ from the centre is $54\mu T$ .What will be its value at the centre of the loop?
A. $250\mu T$
B. $150\mu T$
C. $125\mu T$
D. $75\mu T$
Answer
290.7k+ views
Hint: Relate the given value of magnetic field at the centre of the loop with the formulae for both the cases.
The magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop. For the magnetic field at the centre of the loop, put $z = 0$ in the above expression.
Complete step by step answer:
In the question we are supposed to deal with the magnetic field due to the same circular loop at two different points on the axis and at the centre. Magnetic field on the axis is given ${B_z} = 54\mu T$ .
So we can relate the given value of the magnetic field at the centre of the loop with the formulae for both the cases.
We know that the magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop.
So, let us suppose that current $I$ is flowing in the loop.
Now, for magnetic field at the centre of the loop, we can put $z = 0$ in the above expression which gives
$\implies$ ${B_c} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi I}}{R}$
On dividing both the equation we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}{{{R^3}}}$
Therefore, substituting the values given in the question we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{4^2} + {3^2}} \right)}^{3/2}}}}{{{3^3}}} = \dfrac{{125}}{{27}}$
So, we get the magnetic field at the centre as
$\implies$ ${B_c} = \dfrac{{125}}{{27}} \times {B_z} = \dfrac{{125}}{{27}} \times 54 = 250\mu T$
Hence, option A is correct.
Note: The two formulae which are used here can be derived using Biot-Savart’s Law. The direction of the magnetic field due to the loop can be determined using the right hand thumb rule which states that when we roll our fingers according to the direction of current flowing then the direction of thumb gives us the direction of the magnetic field due to the loop.
The magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop. For the magnetic field at the centre of the loop, put $z = 0$ in the above expression.
Complete step by step answer:
In the question we are supposed to deal with the magnetic field due to the same circular loop at two different points on the axis and at the centre. Magnetic field on the axis is given ${B_z} = 54\mu T$ .
So we can relate the given value of the magnetic field at the centre of the loop with the formulae for both the cases.
We know that the magnetic field on the axis of a current carrying loop is given by ${B_z} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi {R^2}I}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}$ where $z$ is the distance of that point on the axis from centre of the loop, $R$ is the radius of the loop and $I$ is the current flowing in the loop.
So, let us suppose that current $I$ is flowing in the loop.
Now, for magnetic field at the centre of the loop, we can put $z = 0$ in the above expression which gives
$\implies$ ${B_c} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2\pi I}}{R}$
On dividing both the equation we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}{{{R^3}}}$
Therefore, substituting the values given in the question we have
$\implies$ $\dfrac{{{B_c}}}{{{B_z}}} = \dfrac{{{{\left( {{4^2} + {3^2}} \right)}^{3/2}}}}{{{3^3}}} = \dfrac{{125}}{{27}}$
So, we get the magnetic field at the centre as
$\implies$ ${B_c} = \dfrac{{125}}{{27}} \times {B_z} = \dfrac{{125}}{{27}} \times 54 = 250\mu T$
Hence, option A is correct.
Note: The two formulae which are used here can be derived using Biot-Savart’s Law. The direction of the magnetic field due to the loop can be determined using the right hand thumb rule which states that when we roll our fingers according to the direction of current flowing then the direction of thumb gives us the direction of the magnetic field due to the loop.
Last updated date: 02nd Jun 2023
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Total views: 290.7k
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