AP EAPCET 2025 - 21st May Evening Shift | Magnetism and Matter Question 8 | Physics

A short bar magnet is placed in a uniform magnetic field of 2 T such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36 \sqrt{2} \mathrm{~N}-\mathrm{m}$, then the moment of the magnet is

$0.54 \mathrm{JT}^{-1}$

$0.18 \mathrm{JT}^{-1}$

$0.72 \mathrm{JT}^{-1}$

$0.36 \mathrm{JT}^{-1}$

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-0

The relation between $\mu$ and $H$ for a specimen of iron is $\mu=\left[\frac{1.4}{H}+12 \times 10^{-4}\right] \mathrm{Hm}^{-1}$. The value of $H$ which produces flux density of 1 T will be ( $\mu=$ magnetic permeability, $H=$ magnetic intensity)

$250 \mathrm{Am}^{-1}$

$500 \mathrm{Am}^{-1}$

$750 \mathrm{Am}^{-1}$

$10^3 \mathrm{Am}^{-1}$

AP EAPCET 2024 - 21th May Evening Shift

MCQ (Single Correct Answer)

-0

At a place the horizontal component of earth's magnetic field $3 \times 10^{-5} \mathrm{~T}$ and the magnetic declination is $30^{\circ}$. A compass needle of magnetic moment $18 \mathrm{Am}^2$ pointing towards geographic north at this place experiences a torque of

$36 \times 10^{-5} \mathrm{Nm}$

$18 \times 10^{-5} \mathrm{Nm}$

$54 \times 10^{-5} \mathrm{Nm}$

$27 \times 10^{-5} \mathrm{Nm}$

AP EAPCET 2024 - 21th May Morning Shift

MCQ (Single Correct Answer)

-0

If the vertical component of the earth's magnetic field is 0.45 G at a location and angle of dip is $60^{\circ}$, then magnetic field of earth in that location is

0.26 G

0.52 G

0.3 G

0.7 G

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