4. Consider the current loop shown below, consisting of 2 concentric 75 degree a

Question

4. Consider the current loop shown below, consisting of 2 concentric 75 degree arc segments of radii, R1=0.10 m and R2= 0.14 m, connected by 2 segments that are parallel to the center of curvature for the arcs. The current flowing through the loop is 1.5 A, counter-clockwise. Point P is located at the center of curvature for both arc segments. Note:. i. Calculate the magnetic field vector due to the outer segment at point P. ii. Calculate the total magnetic field vector at point P, located at the center of curvature for both arc segments.

Explanation / Answer

here ,

theta = 75 degree ,

R1 = 0.10 m

R2 = 0.14 m

I = 1.5 A

i)

Here , I = 1.5 A

B = (theta/360)*u0*I/(2*R)

B = (75/360)*4pi*10^-7 * 1.5/(2*0.14)

B = (75/360) * 4pi*10^-7 * 1.5/(2 * 0.14)

B = 1.402 *10^-6 T

the magntic field due to outer sagment is 1.402 *10^-6 T

ii)

now, due to both the sagments

B = (75/360)*4pi*10^-7 * 1.5 *(- 1/(2 * 0.14) + 1/(2 * 0.1))

B = 5.61 *10^-7 T

the total magnetic field vector at point P is 5.61 *10^-7 T

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---