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Magnetics question

2018 · 15 Apr · Shift 1 · Q58
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Magnetics question

2018 · 15 Apr · Shift 1 · Q58

JEE MainPhysicsMagneticsMCQ+4 / −1
A Helmholtz coil has a pair of loops, each with NNN turns and radius RRR. They are placed coaxially at distance RRR and the same current I{\rm I}I flows through the loops in the same direction. P,P,P, midway between the centers AAA and CCC, is given by [Refer to figure given below] : JEE Main 2018 (Online) 15th April Morning Slot Physics - Magnetic Effect of Current Question 186 English
  1. A
    8Nμ0I51/2R{{8N{\mu _0}{\rm I}} \over {{5^{1/2}}R}}51/2R8Nμ0​I​
  2. B
    8Nμ0I53/2R{{8N{\mu _0}{\rm I}} \over {{5^{3/2}}R}}53/2R8Nμ0​I​
  3. C
    4Nμ0I51/2R{{4N{\mu _0}{\rm I}} \over {{5^{1/2}}R}}51/2R4Nμ0​I​
  4. D
    4Nμ0I53/2R{{4N{\mu _0}{\rm I}} \over {{5^{3/2}}R}}53/2R4Nμ0​I​
View written solutionFree

Correct answer: B

  1. Magnetic field on the axis of a circular coil

For a circular coil of radius RRR, carrying current III, having NNN turns, the magnetic field at a point on its axis at distance xxx from its center is

B=μ0NIR22(R2+x2)3/2.B = \frac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}.B=2(R2+x2)3/2μ0​NIR2​.
  1. Geometry of Helmholtz coils

In a Helmholtz pair:

  • each coil has radius RRR,
  • the separation between their centers is also RRR,
  • point PPP is midway between the centers.

So the distance of point PPP from the center of each coil is

x=R2.x = \frac{R}{2}.x=2R​.
  1. Field at PPP due to one coil

Using the axial field formula,

B1=μ0NIR22(R2+(R2)2)3/2.B_1 = \frac{\mu_0 N I R^2}{2\left(R^2+\left(\frac{R}{2}\right)^2\right)^{3/2}}.B1​=2(R2+(2R​)2)3/2μ0​NIR2​.

Now simplify:

R2+(R2)2=R2+R24=5R24.R^2+\left(\frac{R}{2}\right)^2 = R^2+\frac{R^2}{4} = \frac{5R^2}{4}.R2+(2R​)2=R2+4R2​=45R2​.

Thus,

(5R24)3/2=53/2R38.\left(\frac{5R^2}{4}\right)^{3/2} = \frac{5^{3/2}R^3}{8}.(45R2​)3/2=853/2R3​.

So,

B1=μ0NIR22⋅53/2R38=μ0NIR253/2R34=4μ0NI53/2R.B_1 = \frac{\mu_0 N I R^2}{2\cdot \frac{5^{3/2}R^3}{8}} = \frac{\mu_0 N I R^2}{\frac{5^{3/2}R^3}{4}} = \frac{4\mu_0 N I}{5^{3/2}R}.B1​=2⋅853/2R3​μ0​NIR2​=453/2R3​μ0​NIR2​=53/2R4μ0​NI​.
  1. Net field at midpoint PPP

Since both currents flow in the same direction, the magnetic fields at the midpoint are in the same direction and add:

B=2B1=2⋅4μ0NI53/2R=8μ0NI53/2R.B = 2B_1 = 2\cdot \frac{4\mu_0 N I}{5^{3/2}R} = \frac{8\mu_0 N I}{5^{3/2}R}.B=2B1​=2⋅53/2R4μ0​NI​=53/2R8μ0​NI​.
  1. Match with options
B=8Nμ0I53/2R\boxed{B = \frac{8N\mu_0 I}{5^{3/2}R}}B=53/2R8Nμ0​I​​

This corresponds to Option B.

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