Question

the torque required to keep a magnet of length 20cm at 30 degress to a uniform field is 2*10^-5n-m.the magnetic force on each pole is

Answer 1

Given that,

Length = 20 cm

Angle = 30°

Torque $\tau= 2\times10^{-5}\ N-m$

We need to calculate the magnetic force on each pole

Using formula of torque

$\tau=F\times x\sin\theta$

$F=\dfrac{\tau}{x\sin\theta}$

$\tau$ = torque

x = perpendicular distance

Put the value into the formula

$F=\dfrac{2\times10^{-5}}{20\sin30\times10^{-2}}$

$F=0.00000338\ N$

$F=3.38\times10^{-6}\ N$

Hence, The magnetic force on each pole is $3.38\times10^{-6}\ N$

Answer 2

Given that,

Length = 20 cm

Angle = 30°

Torque $\tau= 2\times10^{-5}\ N-m$

We need to calculate the magnetic force on each pole

Using formula of torque

$\tau=F\times x\sin\theta$

$F=\dfrac{\tau}{x\sin\theta}$

$\tau$ = torque

x = perpendicular distance

Put the value into the formula

$F=\dfrac{2\times10^{-5}}{20\sin30\times10^{-2}}$

$F=0.00000338\ N$

$F=3.38\times10^{-6}\ N$

Hence, The magnetic force on each pole is $3.38\times10^{-6}\ N$

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