Question

Question 5.8: A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10 −4 m 2 , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10 −2 T is set up at an angle of 30º with the axis of the solenoid?

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-201

Answer 1

(a) number of turns , N = 2000
current , I = 4 A
cross section area , A = 1.6 × 10^-4 m²
we know formula,$\bf{M=NIA}$
so, M = 2000 × 4 × 1.6 × 10^-4
M = 1.28 Am²

(b) force on the solenoid will be zero in uniform magnetic field.
so, torque $\bf{\tau=MBsin\theta}$
= 1.28 × 7.5 × 10^-2 T × sin30°
= 4.8 × 10^-2 Nm
the torque tends to align the solenoid in the direction of magnetic field.

Answer 2

Hey !!

(a) Associated magnetic moment

       μm = niA

        = 2000 × 4 × 1.6 × 10⁻⁴ Am²

       = 1.28 Am²

(b)  Torque = μmBsinθ

             = 1.28 × 7.5 × 10⁻² × sin 30°

             = 0.048 Nm²

GOOD LUCK !!