The magnetic moment of a magnet is 0.1 amp $ \times {{\rm{m}}^2}$. It is suspended in a magnetic field of intensity $3 \times {10^{ - 4}}\;{\rm{Wb}}{{\rm{m}}^{ - 2}}$. The couple acting upon it when deflected by 30$^ \circ $ from the magnetic field is

NEET UG PHYSICS MCQ: MAGNETISM AND MATTER

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The magnetic moment of a magnet is 0.1 amp

\times {\mathrm{m}^2}

. It is suspended in a magnetic field of intensity

3 \times {10^{ - 4}}\;\mathrm{Wb}{\mathrm{m}^{ - 2}}

. The couple acting upon it when deflected by 30

^ \circ

from the magnetic field is

View answer and explanation

Correct option: B

1.5

\times {10^{ - 5}}

N m

Option B is correct because the couple (torque) is calculated using the formula τ = MBsinθ, and substituting the given values yields 1.5 × 10⁻⁵ N m.

Practice All Chapter MCQs →

\vec A = 2\vec i - 3\vec j + 6\vec k\,{m^2}

\vec B = - 3\vec i + \vec j + 2\vec k\,\left( {Wb/{m^2}} \right)

250\;joule

10\;kg

50\;Hz

7.5\;gm/c{m^3}

M

B