A Helmholtz coil has a pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current flows through the loops in the same direction. The magnitude of magnetic field at P, midway between the centres A and C, is given by (Refer to the figure)
A) $\dfrac{{4N{\mu _0}I}}{{{5^{3/2}}R}}$
B) $\dfrac{{8N{\mu _0}I}}{{{5^{3/2}}R}}$
C) $\dfrac{{4N{\mu _0}I}}{{{5^{1/2}}R}}$
D) $\dfrac{{8N{\mu _o}I}}{{{5^{1/2}}R}}$
Answer
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Hint:The area around the magnet where its effect can be experienced, is known as magnetic field. Whenever the charged particles move, they produce a magnetic field. The magnetic field depends on the charge, velocity and the acceleration of the particles.
Complete step-by-step answer:
Step I:
If a coil of wire is placed in a changing magnetic field, then it produces a changing magnetic field. The more the number of turns in the coil, the stronger is the magnetic field.
Step II:
The magnetic field due to both the coils will be equal in magnitude and same in direction. The value of the magnetic field at a point is directly proportional to the value of the current in the conductor and the length of the current-carrying coil.
Step III:
Since the Biot-Savart Law can be used to calculate the strength of the magnetic field at different points. The magnetic field due to one of the circular coil is given by
$B = 2(\dfrac{{{\mu _0}NI{R^2}}}{{2{{({R^2} + \dfrac{{{R^2}}}{2})}^{3/2}}}})$
$B = \dfrac{{4{\mu _0}I{R^2}}}{{{5^{3/2}}{R^3}}}$
Step IV:
Magnetic field due to two coils is given by
${B_{net}} = 2.B$
Substitute the value of B
${B_{net}} = 2 \times \dfrac{{4{\mu _0}I{R^2}}}{{{5^{3/2}}{R^3}}}$
${B_{net}} = \dfrac{{8{\mu _0}I}}{{{5^{3/2}}R}}$
Since there are N turns in the wire and magnetic field is directly proportional to the number of turns
\[{B_{net}} = \dfrac{{8{\mu _0}NI}}{{{5^{3/2}}{R^2}}}\]
Step V:
Option B is the right answer.
Note:It is to be noted that the magnetic field lines due to circular current carrying loops are concentric circles at every point of the loop . The direction of the magnetic field of every section of the circular loop can be found by using the right hand thumb rule. The magnetic field lines are produced in the same direction in the loop. But in the center of the loop, the field lines are straight. The direction of the field lines at the center is perpendicular to the coil.
Complete step-by-step answer:
Step I:
If a coil of wire is placed in a changing magnetic field, then it produces a changing magnetic field. The more the number of turns in the coil, the stronger is the magnetic field.
Step II:
The magnetic field due to both the coils will be equal in magnitude and same in direction. The value of the magnetic field at a point is directly proportional to the value of the current in the conductor and the length of the current-carrying coil.
Step III:
Since the Biot-Savart Law can be used to calculate the strength of the magnetic field at different points. The magnetic field due to one of the circular coil is given by
$B = 2(\dfrac{{{\mu _0}NI{R^2}}}{{2{{({R^2} + \dfrac{{{R^2}}}{2})}^{3/2}}}})$
$B = \dfrac{{4{\mu _0}I{R^2}}}{{{5^{3/2}}{R^3}}}$
Step IV:
Magnetic field due to two coils is given by
${B_{net}} = 2.B$
Substitute the value of B
${B_{net}} = 2 \times \dfrac{{4{\mu _0}I{R^2}}}{{{5^{3/2}}{R^3}}}$
${B_{net}} = \dfrac{{8{\mu _0}I}}{{{5^{3/2}}R}}$
Since there are N turns in the wire and magnetic field is directly proportional to the number of turns
\[{B_{net}} = \dfrac{{8{\mu _0}NI}}{{{5^{3/2}}{R^2}}}\]
Step V:
Option B is the right answer.
Note:It is to be noted that the magnetic field lines due to circular current carrying loops are concentric circles at every point of the loop . The direction of the magnetic field of every section of the circular loop can be found by using the right hand thumb rule. The magnetic field lines are produced in the same direction in the loop. But in the center of the loop, the field lines are straight. The direction of the field lines at the center is perpendicular to the coil.
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