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JEE Main 2018 (15 Apr, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2018 (15 Apr, Shift 1)

Q92JEE Main 2018 · 15 APR · SHIFT 1Difficult

Biot-Savart Law & Circular Coils

A Helmholtz coil has a pair of loops, each with NN turns and radius RR. They are placed coaxially at distance RR and the same current II flows through the loops in the same direction. The magnitude of the magnetic field at P,P, midway between the centers AA and CC, is given by [Refer to figure given below]:

Figure

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Answer: (b)

  1. P is at x=R2x = \dfrac{R}{2} from each coil: B1=μ0NIR22(R2+R24)3/2=μ0NI2R⋅853/2=4μ0NI53/2RB_1 = \dfrac{\mu_0 NIR^2}{2\left(R^2+\frac{R^2}{4}\right)^{3/2}} = \dfrac{\mu_0 NI}{2R}\cdot\dfrac{8}{5^{3/2}} = \dfrac{4\mu_0 NI}{5^{3/2}R}
  2. Same current direction, so the two fields add: B=2B1B = 2B_1.
  3. B=8Nμ0I53/2RB = \dfrac{8N\mu_0 I}{5^{3/2}R}
  4. So the answer is (b).
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