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Magnetic Properties of Matter question

2017 · Shift 0 · Q49
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Magnetic Properties of Matter question

2017 · Shift 0 · Q49

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
A magnetic needle of magnetic moment 6.7 ×\times× 10-2 A m2 and moment of inertia 7.5 ×\times× 10-6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:
  1. A
    8.76 s
  2. B
    6.65 s
  3. C
    8.89 s
  4. D
    6.98 s
View written solutionFree

Correct answer: B

  1. Formula for time period of oscillation of a magnetic needle

For small oscillations in a uniform magnetic field,

T=2πIMBT = 2\pi \sqrt{\frac{I}{MB}}T=2πMBI​​

where:

  • I=7.5×10−6 kg m2I = 7.5 \times 10^{-6}\,\text{kg m}^2I=7.5×10−6kg m2
  • M=6.7×10−2 A m2M = 6.7 \times 10^{-2}\,\text{A m}^2M=6.7×10−2A m2
  • B=0.01 TB = 0.01\,\text{T}B=0.01T
  1. Compute MBMBMB

MB=(6.7×10−2)(0.01)=6.7×10−4MB = (6.7 \times 10^{-2})(0.01) = 6.7 \times 10^{-4}MB=(6.7×10−2)(0.01)=6.7×10−4

  1. Compute the ratio IMB\frac{I}{MB}MBI​

IMB=7.5×10−66.7×10−4\frac{I}{MB} = \frac{7.5 \times 10^{-6}}{6.7 \times 10^{-4}}MBI​=6.7×10−47.5×10−6​

=7.56.7×10−2≈1.1194×10−2= \frac{7.5}{6.7} \times 10^{-2} \approx 1.1194 \times 10^{-2}=6.77.5​×10−2≈1.1194×10−2

  1. Take square root

IMB=1.1194×10−2≈0.1058\sqrt{\frac{I}{MB}} = \sqrt{1.1194 \times 10^{-2}} \approx 0.1058MBI​​=1.1194×10−2​≈0.1058

  1. Find time period for one oscillation

T=2π(0.1058)≈6.2832×0.1058≈0.665 sT = 2\pi(0.1058) \approx 6.2832 \times 0.1058 \approx 0.665\,\text{s}T=2π(0.1058)≈6.2832×0.1058≈0.665s

  1. Time for 10 oscillations

t=10T=10×0.665=6.65 st = 10T = 10 \times 0.665 = 6.65\,\text{s}t=10T=10×0.665=6.65s

  1. Match with options

6.65 s\boxed{6.65\,\text{s}}6.65s​

So, the correct option is B.

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