A circular loop of radius12cmcarries a current of15A. A flat coil of radius0.82cmhaving50turnsand a current of 1.3A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop’s magnetic field is uniform across the coil. (a) What is the magnitude of the magnetic field produced by the loop at its center and (b) What is the magnitude of the torque on the coil due to the loop?

Short Answer

Expert verified

(a) Magnetic field produced at the center is 7.9×105T.

(b) Torque on coil due to loop is 1.1×106Nm.

Step by step solution

01

Identification of the given data

  • The loop radius,r=12 cm
  • The current, i=15 A
  • The flat coil radius, R=0.82cm
  • The number of turns, N=50
  • The current through coil, I=1.3 A
02

Understanding the concept:

Define the magnetic field due to the circular loop at center by using the formula in terms of current and radius. Then, you use this magnetic field to find torque using the relation between torque, magnetic field, and magnetic dipole moment.

Formulae:

The magnetic field is defined by,

B=μ0i2r

The magnetic dipole moment is,

μ=NiA

Here, μ0is the permeability of free space having a value 4π×107TmA,iis the current,Iis the current through coil,ris the radius, Bis the magnetic field, Nis the number of turns, and Ais the area.

03

(a) Calculate magnetic field produced at the center:

Define the magnetic field at center as follows.

B=μi2r

Substitute all the value in the above equation.

B=4π×107Tm/A×15A2×0.12m=7.9×105T

Hence, the magnetic field is role="math" localid="1663248173578" 7.9×105T.

04

(b) Calculate Torque on coil due to loop:

Now torque acting on coil of current Idue to the loop with magnetic field Bis as follows.

τ=NIABsinθ=N×I×π×R2×B×sin90°

Substitute all the value in the above equation.

τ=50×1.3A×3.14×(0.0082m)2×(7.9×105T)×1=1.1×106Nm

Hence, the torque is 1.1×106Nm.

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