The rate of change of torque $τ$ with deflection $θ$ is maximum for a magnet suspended freely in a uniform magnetic field of induction B when $θ$ is equal to

0^{o}

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45^{o}

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60^{o}

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90^{o}

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Solution

The correct option is A $0^{o}$
$\frac{d z}{d t} = ϵ$
$z = m B sin θ$
$\frac{d z}{d t} = m B cos θ \frac{d θ}{d t}$
$F o r (\frac{d z}{d t}) m a x$
$\frac{d^{2} z}{d f^{2}} = 0$
$\frac{d^{2} z}{d f^{2}} = m B sin θ \frac{d^{2} θ}{d f^{2}} = 0$
$sin θ = 0$
$θ = 0$

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The rate of change of torque τ with deflection θ is maximum for a magnet suspended freely in a uniform magnetic field of induction B when θ is equal to

The correct option is A 0odzdt=ϵz=mBsinθdzdt=mBcosθdθdtFor(dzdt)maxd2zdf2=0d2zdf2=mBsinθd2θdf2=0sinθ=0θ=0

The rate of change of torque $τ$ with deflection $θ$ is maximum for a magnet suspended freely in a uniform magnetic field of induction B when $θ$ is equal to

The correct option is A $0^{o}$
$\frac{d z}{d t} = ϵ$
$z = m B sin θ$
$\frac{d z}{d t} = m B cos θ \frac{d θ}{d t}$
$F o r (\frac{d z}{d t}) m a x$
$\frac{d^{2} z}{d f^{2}} = 0$
$\frac{d^{2} z}{d f^{2}} = m B sin θ \frac{d^{2} θ}{d f^{2}} = 0$
$sin θ = 0$
$θ = 0$