A bar magnet of length 0.2 m and pole strength 5 A.m. should be kept in a uniform magnetic field of induction 15 tesla at angle ..... radians to the field so that the torque experienced by it will be 7.5N-m

\frac{π}{2}

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\frac{π}{3}

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\frac{π}{6}

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\frac{π}{4}

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Solution

The correct option is C $\frac{π}{6}$
$l = 0.2$
$P = 50 m$
$B = 15 T$
$z = 7.5 N m$
$\to z = \to m \times \to B m = l P$
$z = m B s i n θ$
$7.5 = 0.2 \times 5 \times 15 \times s i n θ$
$s i n θ = \frac{1}{2}$
$θ = π / 6$

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A bar magnet of length 0.2 m and pole strength 5 A.m. should be kept in a uniform magnetic field of induction 15 tesla at angle ..... radians to the field so that the torque experienced by it will be 7.5N-m

The correct option is C π6l=0.2P=50mB=15Tz=7.5Nm→z=→m×→Bm=lPz=mBsinθ7.5=0.2×5×15×sinθsinθ=12θ=π/6

The correct option is C $\frac{π}{6}$
$l = 0.2$
$P = 50 m$
$B = 15 T$
$z = 7.5 N m$
$\to z = \to m \times \to B m = l P$
$z = m B s i n θ$
$7.5 = 0.2 \times 5 \times 15 \times s i n θ$
$s i n θ = \frac{1}{2}$
$θ = π / 6$