/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q15E Lightning bolts can carry curren... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Lightning bolts can carry currents up to approximately 20 kA . We can model such a current as the equivalent of a very long, straight wire. (a) If you were unfortunate enough to be 5m away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being 5cm from a long, straight household current of 10A ?

Short Answer

Expert verified

a)B=8×10-4T

b)B=4×10-5T

Step by step solution

01

Take lightening as a very long wire

To simplify the calculations, we will take lightening as a very long wire. The formula of magnetic field will becomeB=μ0I2Ï€°ù .

Given thatI=20kA and r = 5m .

B=μ0I2πrB=4π×10−7×200002π×5B=8×10−4T

02

Calculate the magnetic field for the second case

In the second case, r = 5 cm and I = 10 A .

B=4π×10−7×102π×5×10−2B=4×10−5T

We can see that field due to lightening bolt is about twenty times as strong as the field due to the household.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel-plate capacitor is charged by being connected To a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric Field change? The potential difference? The total energy? Explain.

Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00 nC is on the x-axis at x = 3.00 cm. Point P is on the y-axis at y = 4.00 cm. (a) Calculate the electric fieldsandat point P due to the charges q1 and q2. Express your results in terms of unit vectors (see Example 21.6). (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

Two identical metal objects are mounted on insulating stands. Describe how you could place charges of opposite sign but exactly equal magnitude on the two objects.

When a resistor with resistance Ris connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that Rremains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

A very small sphere with positive charge q=+8.0μCis released from rest at a point 1.50cmfrom a very long line of uniform linear charge density λ=+3.00μC/m. What is the kinetic energy of the sphere when it is 4.50cmfrom the line of charge if the only force on it is the force exerted by the line of charge?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.