A bar magnet of length $10 c m$ and pole strength $2 A m$ makes an angle $60^{o}$ with a uniform magnetic field of induction $50 T$ . The couple acting on it :

5 \sqrt{3} N m

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\sqrt{3} N m

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10 \sqrt{3} N m

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20 \sqrt{3} N m

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Solution

The correct option is A $5 \sqrt{3} N m$
Given,
Length, $l = 10 c m$

Pole strength, $P = 2 A m$

Angle, $θ = 60$

Magnetic field induction $B = 50 T$

$m = l P$

Couple, $z = m B sin θ$

$= \frac{10}{100} \times 2 \times 50 \times \frac{\sqrt{3}}{2}$

$= 5 \sqrt{3} N m$

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A bar magnet of length 10 cm and pole strength 2 Am makes an angle 60o with a uniform magnetic field of induction 50T. The couple acting on it :

The correct option is A 5√3NmGiven, Length, l=10 cmPole strength, P=2 AmAngle, θ=60Magnetic field inductionB=50Tm=lPCouple, z=mBsinθ=10100×2×50×√32=5√3Nm

A bar magnet of length $10 c m$ and pole strength $2 A m$ makes an angle $60^{o}$ with a uniform magnetic field of induction $50 T$ . The couple acting on it :

The correct option is A $5 \sqrt{3} N m$
Given,
Length, $l = 10 c m$

Pole strength, $P = 2 A m$

Angle, $θ = 60$

Magnetic field induction $B = 50 T$

$m = l P$

Couple, $z = m B sin θ$

$= \frac{10}{100} \times 2 \times 50 \times \frac{\sqrt{3}}{2}$

$= 5 \sqrt{3} N m$