Question

Scientists have created an electric field of magnitude E_(A) in the z-y plane at an angle \theta in a region of space, and in the same region have created a magnetic field of magnitude B_(A) oriented as shown. A spaceship travels in the x-direction at speed vv_(BA). Consider the case where E_(A)=3.2\times 10^(8)(N)/(()/())C,B_(A)=0.84T,v_(BA)=2.2\times 10^(7)(m)/(s), and \theta =24\deg . What is the magnitude of the electric field as seen by the spaceship? Enter your answer in M(N)/(C).

          Scientists have created an electric
field of magnitude E_(A) in the z-y
plane at an angle \theta  in a region of
space, and in the same region have
created a magnetic field of
magnitude B_(A) oriented as shown. A
spaceship travels in the x-direction
at speed vv_(BA).
Consider the case where E_(A)=3.2\times 
10^(8)(N)/(()/())C,B_(A)=0.84T,v_(BA)=2.2\times 
10^(7)(m)/(s), and \theta =24\deg .
What is the magnitude of the
electric field as seen by the
spaceship?
Enter your answer in M(N)/(C).

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scientists have created an electric field of magnitude ea in the z y plane at an angle theta in a region of space and in the same region have created a magnetic field of magnitude ba orien 21947

Added by Jason B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Suzanne W.

04/26/2022

Step 1

A spaceship travels in the x-direction at speed $$v_{BA}$$. Consider the case where $$E_A = 3.2 \times 10^8 N/C$$, $$B_A = 0.84 T$$, $$v_{BA} = 2.2 \times 10^7 m/s$$, and $$\theta = 24^\circ$$. What is the magnitude of the electric field as seen by the Show more…

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Scientists have created an electric field of magnitude E_(A) in the z-y plane at an angle \theta in a region of space, and in the same region have created a magnetic field of magnitude B_(A) oriented as shown. A spaceship travels in the x-direction at speed vv_(BA). Consider the case where E_(A)=3.2\times 10^(8)(N)/(()/())C,B_(A)=0.84T,v_(BA)=2.2\times 10^(7)(m)/(s), and \theta =24\deg . What is the magnitude of the electric field as seen by the spaceship? Enter your answer in M(N)/(C).

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00:01 Okay, so scientists have created electric and magnetic fields as shown in the figure from your problem.

00:08 These fields are also seen by scientists that zoom pass in a rocket traveling in the x direction.

00:15 So the velocity in the x direction is 1 times 10 to the 6th in units of meters per second.

00:23 We know that the magnitude of the magnetic field is 0 .48 tesla and the magnitude of the electric field is 1 .2, 10 to the 6 in units of volts per meter.

00:38 And we need to find the angle that it's going to make.

00:43 So start with the net electric field.

00:47 E net is equal to e plus v cross b in the j hat direction, since you have a magnetic field.

00:59 So e plus v cross b.

01:04 This is e cosine of theta in the i -hat direction, minus e sine of theta in the j -hat direction.

01:13 So that's just the components of e.

01:15 And then v -cross -b is v times b in the j -hat direction.

01:24 They told you that the angle is 45 given from the figure.

01:28 So i'm just going to go ahead and sub that in.

01:32 45 degrees, 45 degrees.

01:39 And then let's combine components.

01:42 So this is e cosine of 45 degrees in the i -hat direction.

01:49 And then vb minus e sign of 45 degrees.

01:54 All of this is in the j -hat direction.

01:57 We need that because theta is equal to the inverse tangent of whatever is in the y direction over whatever is in the x direction...

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