Question

3) There is a current of 0.2 A in the circuit consisting of a ring-shaped wire with a radius of 8 cm. The vector of unit length and parallel to the μ ⃗ dipole moment of the circuit is given as 0.6i - 0.8j (This unit shows the direction of the vector magnetic dipole moment). If this circuit is in a uniform magnetic field given by B ⃗ ⃗ = (0.25i + 0.3k ⃗) T

a) Torque on the circuit (with unit vector representation) and

b) Find the directional energy of the circuit

Answer 1

Wam] 3 GIVEN: 1. Current in the circuit = 0.2A 2.) Radius of ving = 8cm 3.) Unit vector along 4) magnetic field (18) = (0.25 +0.35) Solution : along the dipole moment = 0.65 -0.8; PART-| First of all of will have to calculate magnetic moment Circuit (R) and formula to Calculate Min M is given as hehere is current in the wire Area vector A - .: M. IR = 0.2 (101) magnitude of Arca 0 direction of Avea vector that will be along dipole - 0.2 (xx008 x008 moment. 0.650.87) 0.004 (0:61 - 0P1) = 0.0024 T-0.00321 Now, we know that torque acting on Circuit will be: Z-My B n = magnetic [where B = magnetic (0.00245 -0.00325) field moment X (0.251 tot

Ka 0.0024 -0.0032 0 0.25 003 0.00324 0.3 -0.0024x03 +F Co 0 +0.0032. X025 + 84104 104 = -96X184 - 12.2004 (-3472 9.6ł - 7.25 +86 ) Nom Hence, Required Torque ** (-3,61 -221 +82) PART-2 on the Circuit N-m b. Directional Energy is given by the formula: Directional Energy = -0.13 hehere, n = magnetic moment B = Magnetic field Directional Energy = -9.13 0.0024 24 1-0 -0.00325 tol [ odsł toj tosi - [oooo 0.0024 x 0.25] Joule, = =-68109 Joule,