Use the following to answer the next two questions: -A coil of radius R = 5 cm lies inside a long solenoid which carries a current I₁ = 10 A and has 800 turns/m.The coil carries a current I₂ = 4 A and the normal of the plane of the coil is at an angle $\theta$ = 30 $\circ$ with the axis of the solenoid.The magnetic field produced by the solenoid at the center is ____ times greater than the magnetic field by the coil at the center. A)100 B)200 C)1000 D)2000 E)8000

Question

Quizplus · Accepted Answer

The magnetic field produced by the solenoid at the center is given by:
Bsol = μ0*N*IS/l
where μ0 is the magnetic constant, N is the number of turns per unit length (800 turns/m), IS is the current through the solenoid (10 A) and l is the length of the solenoid.

The magnetic field produced by the coil at the center is given by:
Bcoil = μ0*IS*(πR^2)/d^3
where IS is the current through the coil (4 A), R is the radius of the coil (5 cm), d is the distance from the center of the coil to the center of the solenoid, and μ0 is the magnetic constant.

To calculate the ratio of Bsol/Bcoil, we need to find the value of Bsol and Bcoil at the center of the solenoid. Since the coil is located at the center of the solenoid, we can assume that d is equal to the radius of the solenoid.

Substituting the given values, we get:
Bsol = (4π*10^-7)*800*10/(0.05) = 2.54 T
Bcoil = (4π*10^-7)*4*(π*0.05^2)/(0.05^3) = 0.5 T

Therefore, the ratio of Bsol/Bcoil is:
Bsol/Bcoil = 2.54/0.5 = 5.08 = 200% (approximately)

Hence, the answer is B, 200.

Use the Following to Answer the Next Two Questions θ\thetaθ = 30

Use the Following to Answer the Next Two Questions $\theta$ = 30