Question

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DO 4 Out 01 . Only 4 will be graded. Cross out the one you don't do. +10 C moves at 20 m/s in the LA particle with charge positive r-direction (see diagram). There is a magnetic field of L9 x10 T out of the page. What is the force on the particle? (Give magnitude and direction!) 引0(20) (1 AXIO-3 TO F = 0.38 -regohie in the ydirection The subsequent motion of this particle is a circular path. Is it clockwise or counterclockwise? b. c. If the particle's mass is 1.5 x 10s kg, what is the radius of the particle's cyclotron motion? -5 -1 2. You are standing near a high voltage power line and detect a magnetic field around it. a. Sketch the magnetic field that will be formed around a current-carrying wire.y Neatness counts 个 b. You detect a field of 4 HT when you are precisely 13 m away. What is the current in the wire? т" 3.BH o KI0""A c. The Earth's magnetic field strength is around 5 x 10-S T on the Earth's surface. How far away from the wire would you have to be for the wire's magnetic field to exactly equal the Earth's magnetic field in magnitude? ser beam strikes the surface of a strange liquid at an angle of 32° with the normal. The refracted beam makes an ni angle of 19° with the normal, in the liquid. 02 a. What's the index of refraction for the strange liquid? sMi 뽑흙 "BurH31ー1 @y큼 음 b. Estimate the speed of light in this liquid. Seems 201 slow c. What's the critical angle for light shining out of this liquid? 9031.00 see a ladybug (2 cm from the lens) upside-down, three times bigger than real life. 4. You're using a convex lens as a magnifying glass, and you a. Find d. 2 b. Find f c. Draw a picture showing (to scale) the location of the lens, the ladybug, and the image. Neatness counts.

Answer 1

(1)

Force on a charge particle having charge 'q' , moving with a velocity 'v' in a magnetic field of strength 'B' is given by

  $\vec{F} = q (\vec{v} \times \vec{B})$

Given, q = 10 C , v = 20 m/s (along +ve x-axis) and B = 1.9 x 10^-3 T out from the page (or in positive Z direction)

In vector notation, we can write

  $\vec{v} =20 \,\, \hat{i}$ and

  

B = 1.9 × 10-3 k

(a) Therefore, the force on the charge particle is

   in N

F-10 × (20 ^ × 1.9 × 10-3 k) 0.380 (j)

Hence, the direction of force is along the negative y-axis.

(b) As the force is along the negative y-axis, the particle will move in clockwise direction.

(c) The radius of the circular path of the charge particle moving in a magnetic field is given by

1.5 × 10-8 × 20 10 × 1.9 × 10-3 mv 1.58 × 10-5 m 4B

(2) Magnetic field due to a current conducting wire at a distance 'r ' is given by

μοι

Where, I is the current in the wire

(a) The direction of the field lines across a wire is decided by the "right hand thumb rule".

The arrow is denoting the direction of magnetic field across the wire.

(b) Given at a distance of 13 m, the field strength was 4 x 10^-6 T. Then we can write,

4π× 10-1 ×1 2π × 13 4 × 10-o-

or,   

0.2 × 13

or,

13x4 = 260 A 0.2

Hence, the wire has a current of 260 A.

(c) Given, Earth's magnetic field is 5 x 10^-5 T .

Suppose we have to be ' r ' distance away from the wire for the wire's magnetic field to be exactly equals to the Earth's field. Then we can write

  $5 \times 10^{-5} = \frac{4 \pi \times 10^{-7} \times 260}{2 \pi \times r}$

or,  $5 = \frac{2 \times 10^{-2} \times 260}{ r}$

or,  $r = \frac{2 \times 10^{-2} \times 260}{ 5}$

or,  

Do you know, asking more than one question at a time is against the policy. still i have managed to answer the first two questions. Please upload other two separately.

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