A square loop (side a) is mounted on a vertical shaft and rotated at angular velocity ω (Fig. 7.19). A uniform magnetic field B points to the right. Find theεtfor this alternating current generator.

Short Answer

Expert verified

The induced emf in the square loop is .Ba2ωsinωt

Step by step solution

01

Write the given data from the question.

The uniform magnetic field is B .

The side of the square loop is a .

The angular velocity is ω.

02

Calculate the generated emf in the square loop.

εt=-ddt(BAcosωt)The area of square loop,A=a2 .

The square loop is moving at angle with angular velocity in time

tθωt.

The magnetic flus through the loop is given by,

role="math" localid="1657618600560" ϕ=BAcosθ

Substitute ωtfor θinto above equation.

ϕ=BAcosθ

According to the Faraday’s law, the induced emf in any closed loop is equal to the negative of the rate of change of flux in the circuit.

ε(t)=-dϕdt

Substitute ϕ=BAcosωtfor ϕinto above equation.

εt=-ddt(BAcosωt)

Substitute for into above equation.

role="math" localid="1657619049126" εt=-ddt(BAcosωt)εt=-Ba2-sinωt.ωεt=-Ba2sinωt

Hence the induced emf in the square loop is Ba2ωsinωt.

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Most popular questions from this chapter

Suppose a magnetic monopole qm passes through a resistanceless loop of wire with self-inductance L . What current is induced in the loop?

A square loop of wire (side a) lies on a table, a distance s from a very long straight wire, which carries a current I, as shown in Fig. 7.18.

(a) Find the flux of B through the loop.

(b) If someone now pulls the loop directly away from the wire, at speed, V what emf is generated? In what direction (clockwise or counter clockwise) does the current flow?

(c) What if the loop is pulled to the right at speed V ?

An infinite wire runs along the z axis; it carries a current I (z) that is a function ofz(but not of t ), and a charge density λ(t) that is a function of t (but not of z ).

(a) By examining the charge flowing into a segment dz in a time dt, show that dλ/dt=-di/dz. If we stipulate that λ(0)=0and I(0)=0, show that λ(t)=kt, I(z)=-kz, where k is a constant.

(b) Assume for a moment that the process is quasistatic, so the fields are given by Eqs. 2.9 and 5.38. Show that these are in fact the exact fields, by confirming that all four of Maxwell's equations are satisfied. (First do it in differential form, for the region s > 0, then in integral form for the appropriate Gaussian cylinder/Amperian loop straddling the axis.)

A familiar demonstration of superconductivity (Prob. 7.44) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. Treat the magnet as a perfect dipole , m a height z above the origin (and constrained to point in the z direction), and pretend that the superconductor occupies the entire half-space below the xy plane. Because of the Meissner effect, B = 0 for Z0, and since B is divergenceless, the normal ( z) component is continuous, so Bz=0just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at - z , as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region z>0.

(a) Which way should the image dipole point (+ z or -z)?

(b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to Mg (where M is the mass of the magnet) to determine the height h at which the magnet will "float." [Hint: Refer to Prob. 6.3.]

(c) The induced current on the surface of the superconductor ( xy the plane) can be determined from the boundary condition on the tangential component of B (Eq. 5.76): B=μ0(K×z^). Using the field you get from the image configuration, show that

K=-3mrh2π(r2+h2)52ϕ^

where r is the distance from the origin.

Question:Calculate the energy stored in the toroidal coil of Ex. 7.11, by applying Eq. 7.35. Use the answer to check Eq. 7.28.

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