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Along the path shown in Figure 21.46 the magnetic field is measured and is found to be uniform in magnitude and always tangent to the circular path. If the radius of the path is 3cmand Balong the path is 1.3×10-6T, what are magnitude and direction of the current enclosed by the path?

Short Answer

Expert verified

The magnitude of the current enclose by path is 0.195Aand the direction is into the plane of the paper

Step by step solution

01

Identification of given data

Strength of the magnetic fieldB=1.3×10-6T

Radius of the pathR=3cmor0.03m

02

Significance of Ampere's circuital equation

According to Ampere's circuital equation, the line integral of the magnetic field forming a closed loop around a current-carrying wire in a plane normal to the current is equal to μotimes the net current flowing through the loop.

B·dL=μoI ...(i)

Where, Bis the strength of the magnetic field, dLis the length of elementary part, μois the permeability of the medium (4π×10-7H/m) andIis the strength of current

03

Determining the magnitude and direction of the current enclosed by the path

By using equation (i)

B·dL=BdL=B2πR

B2πR=μoII=B×2πRμo

Substitute all the values in above equation

I=1.3×10-6T×2π×0.03m4π×10-7H/m=0.195A

Hence the magnitude of the current enclose by path is 0.195Aand the direction is into the plane of the paper

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Most popular questions from this chapter

Figure 21.70 shows a close-up of the central region of a capacitor made of two large metal plates of area , very close together and charged equally and oppositely. There are +Q and -Q on the inner surfaces of the plates and small amounts of charge +q and -q on the outer surfaces.

(a). Knowing Q, determine E: Consider a Gaussian surface in the shape of a shoe box, with one end of area Abox in the interior of the left plate and the other end in the air gap (surface 1 in the diagram). Using only the fact that the electric field is expected to be horizontal everywhere in this region, use Gauss’s law to determine the magnitude of the field in the air gap. Check that your result agrees with our earlier calculations in the Chapter 15. (Be sure to consider the flux on all faces of your Gaussian box.)

A lead nucleus is spherical with a radius of about 7×10-15m . The nucleus contains 82protons (and typically126 neutrons). Because of their motions the protons can be considered on average to be uniformly distributed throughout the nucleus. Base on the net flux at the surface of the nucleus, calculate the divergence of the electric field as electric flux per unit volume. Repeat the calculation at a radius of3.5×10-15m . (You can use Gauss’s law to determine the magnitude of the electric field at this radius.) Also calculate the quantityρ/ε0 inside the nucleus.

Figure 21.62 shows a box on who surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of electric field is 400 V/m and on the right face the magnitude of electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux and total amount of charge that is inside the box.

The electric field has been measured to be horizontal and to the right everywhere on the closed box as shown in Figure 21.66. All over the left side of box E1=100V/m and all over the right (slanting) side of box E2=300V/m.On the top the average field is E3=150V/m , on the front and back the average field is E4=175V/mand on the bottom the average field is E5=220V/m.How much charge is inside the box? Explain briefly.

A straight circular plastic cylinder of length L and radius R ( where R<<L) is irradiated with a bean of protons so that there is a total excess charge Q distributed uniformly throughout the cylinder. Find the electric field inside the cylinder, a distance r from the center of the cylinder far from the ends, where r < R.

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