JEE MAIN - Physics (2017 (Offline) - No. 6)
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:
8.76 s
6.65 s
8.89 s
6.98 s
Explanation
Given : Magnetic moment, M = 6.7 × 10–2 Am2
Magnetic field, B = 0.01 T
Moment of inertia, I = 7.5 × 10–6 Kgm2
Using, T = $$2\pi \sqrt {{I \over {MB}}} $$
= $$2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}} $$
= $${{2\pi } \over {10}} \times 1.06$$ s
Time taken for 10 complete oscillations
t = 10T = 2$$\pi $$ × 1.06
= 6.6568 $$ \simeq $$ 6.65 s
Magnetic field, B = 0.01 T
Moment of inertia, I = 7.5 × 10–6 Kgm2
Using, T = $$2\pi \sqrt {{I \over {MB}}} $$
= $$2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}} $$
= $${{2\pi } \over {10}} \times 1.06$$ s
Time taken for 10 complete oscillations
t = 10T = 2$$\pi $$ × 1.06
= 6.6568 $$ \simeq $$ 6.65 s
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