A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10-4 m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10-2 tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be

Question

A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10-4 m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10-2 tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be (A) 3 × 10-3 N m (B) 1.5 × 10-3 N m (C) 1.5 × 10-2 Nm (D) 3 × 10-2 Nm

Tardigrade Admin · Accepted Answer

Magnetic moment of the loop. M=N I A=2000 × 2 × 1.5 × 10-4=0.6 J / T Torque τ= MB sin ° 30° =0.6 × 5 × 10-2 × (1/2)=1.5 × 10-2 N m

$3 \times 1 0^{- 3} N m$

$1.5 \times 1 0^{- 3} N m$

$1.5 \times 1 0^{- 2} N m$

$3 \times 1 0^{- 2} N m$

Magnetic moment of the loop.
$M = N I A = 2000 \times 2 \times 1.5 \times 1 0^{- 4} = 0.6 J / T$
Torque $τ = MB sin^{\circ} 3 0^{\circ}$
$= 0.6 \times 5 \times 1 0^{- 2} \times \frac{1}{2} = 1.5 \times 1 0^{- 2} N m$

3×10−3Nm

1.5×10−3Nm

1.5×10−2Nm

3×10−2Nm

Magnetic moment of the loop. M=NIA=2000×2×1.5×10−4=0.6J/T Torque τ=MBsin∘30∘ =0.6×5×10−2×21​=1.5×10−2Nm

$3 \times 1 0^{- 3} N m$

$1.5 \times 1 0^{- 3} N m$

$1.5 \times 1 0^{- 2} N m$

$3 \times 1 0^{- 2} N m$

Magnetic moment of the loop.
$M = N I A = 2000 \times 2 \times 1.5 \times 1 0^{- 4} = 0.6 J / T$
Torque $τ = MB sin^{\circ} 3 0^{\circ}$
$= 0.6 \times 5 \times 1 0^{- 2} \times \frac{1}{2} = 1.5 \times 1 0^{- 2} N m$