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Byju's Answer

Standard XII

Physics

Rotational EMF

A bar magnet ...

Question

A bar magnet of length $10 c m$ experiences a torque of $\frac{\sqrt{2}}{10} N m$ in an uniform magnetic field of induction $0.4 W b / m^{2}$ when it is suspended making an angle $45^{o}$ with the field. The pole strength of the magnet is ( in $A m$ ) :

50

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10

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2.5

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5

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Solution

The correct option is D $5$
The torque on the magnet $= \to M \times \to B = m l B sin θ = \frac{\sqrt{2}}{10} N m$

$\Rightarrow m \times 0.1 \times 0.4 \times sin 45^{0} = \frac{\sqrt{2}}{10}$
$m = 5 c m$
Therefore, D is the correct option

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A bar magnet of length 10cm experiences a torque of √210Nm in an uniform magnetic field of induction 0.4Wb/m2 when it is suspended making an angle 45o with the field. The pole strength of the magnet is ( in Am) :

The correct option is D 5The torque on the magnet =→M×→B=mlBsinθ=√210Nm⇒m×0.1×0.4×sin450=√210m=5cmTherefore, D is the correct option

A bar magnet of length $10 c m$ experiences a torque of $\frac{\sqrt{2}}{10} N m$ in an uniform magnetic field of induction $0.4 W b / m^{2}$ when it is suspended making an angle $45^{o}$ with the field. The pole strength of the magnet is ( in $A m$ ) :

The correct option is D $5$
The torque on the magnet $= \to M \times \to B = m l B sin θ = \frac{\sqrt{2}}{10} N m$

$\Rightarrow m \times 0.1 \times 0.4 \times sin 45^{0} = \frac{\sqrt{2}}{10}$
$m = 5 c m$
Therefore, D is the correct option