A closely wound solenoid of 2000 turns and area of cross-section1.6 × 10–4 m2, carrying a current of 4.0 A, is suspended through itscentre 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 uniformhorizontal magnetic field of 7.5 × 10–2 T is set up at an angle of30° with the axis of the solenoid?
Given: Number of turns in the solenoid is 2000 , cross-sectional area of the solenoid is 1.6 × 10 − 4 m 2 and the current passing through the solenoid is 4 A .
a)
The magnetic moment along the axis of solenoid is given as,
m = N I A
Where, N is the number of turns, I is the current passing through the solenoid and A is the area of the solenoid.
By substituting the given values in the above equation, we get
m = 2000 × 4 × 1.6 × 10 − 4 = 1.28 A-m 2
The direction of this magnetic moment will be along the axis of the solenoid in the direction of the current (Using right hand screw rule to determine the direction of magnetic moment.).
Thus, the magnetic moment associated with the solenoid is 1.28 A-m 2 .
b)
Given: The magnetic field is 7.5 × 10 − 2 T and angle between axis of solenoid and magnetic field is 30 ° .
The torque acting on a solenoid is given as,
τ = m B sin θ
Where, B is the magnetic field and θ is the angle between magnetic moment and magnetic field of the solenoid.
By substituting the given values in the above equation, we get
τ = 1.28 × 7.5 × 10 − 2 × sin 30 ° = 1.28 × 7.5 × 10 − 2 × 0.5 = 4.8 × 10 − 2 Nm
The force acting on a solenoid is given as,
F → = I L → × B →
Where, F is the force acting on the solenoid, I is the current passing through the solenoid and L is the length of the solenoid.
Since, magnetic field is uniform throughout the solenoid, so the force at any small element of the solenoid comes out to be radially outweard and when added vectorially it will be zero. Thus, the force acting on the solenoid is 0 .
Thus, the torque and force on the solenoid is 4.8 × 10 − 2 Nm and zero respectively.
Given: Number of turns in the solenoid is
a)
The magnetic moment along the axis of solenoid is given as,
Where,
By substituting the given values in the above equation, we get
The direction of this magnetic moment will be along the axis of the solenoid in the direction of the current (Using right hand screw rule to determine the direction of magnetic moment.).
Thus, the magnetic moment associated with the solenoid is
b)
Given: The magnetic field is
The torque acting on a solenoid is given as,
Where,
By substituting the given values in the above equation, we get
The force acting on a solenoid is given as,
Where,
Since, magnetic field is uniform throughout the solenoid, so the force at any small element of the solenoid comes out to be radially outweard and when added vectorially it will be zero. Thus, the force acting on the solenoid is
Thus, the torque and force on the solenoid is
