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JEE Main 202224 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2022 · 24 June · Morning Shift · Physics · Magnetic Effects of Current

The magnetic field at the centre of a circular coil of radius r ,...

The magnetic field at the centre of a circular coil of radius rr, due to current I flowing through it, is BB. The magnetic field at a point along the axis at a distance r2\dfrac{r}{2} from the centre is :

Options

  1. AB2\dfrac{B}{2}
  2. B2B2B
  3. C(25)3B{\left(\dfrac{2}{\sqrt{5}}\right)}^{3}B
  4. D(23)3B{\left(\dfrac{2}{\sqrt{3}}\right)}^{3}B

Correct answer

C. (25)3B{\left(\dfrac{2}{\sqrt{5}}\right)}^{3}B

Step-by-step solution

Magnetic field at the centre of the circle is, B=μ0i2RB = \dfrac{{\mu}_{0}i}{2R}

Magnetic field at the point on the axis is,

B2=μ0iR22(R2+x2)32⇒B2=μ0iR22(R2+(R2)2)32⇒B2=μ0i2(54)32R=(25)3B{B}_{2} = \dfrac{{\mu}_{0}i{R}^{2}}{2{\left({R}^{2} + {x}^{2}\right)}^{\dfrac{3}{2}}} ⇒ {B}_{2} = \dfrac{{\mu}_{0}i{R}^{2}}{2{\left({R}^{2} + {\left(\dfrac{R}{2}\right)}^{2}\right)}^{\dfrac{3}{2}}} ⇒ {B}_{2} = \dfrac{{\mu}_{0}i}{2{\left(\dfrac{5}{4}\right)}^{\dfrac{3}{2}}R} = {\left(\dfrac{2}{\sqrt{5}}\right)}^{3}B

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