Question #b27b7

2 Answers
Sep 9, 2017

#"a)"# #1.25 times 10^(- 13)# #"N"#

#"b)"# #2.89 times 10^(- 2)# #"N/m"#

Explanation:

#"1"#

#"a)"#

We are given two vectors, one for the velocity #(vec(v))# and one for the magnetic field #(vec(B))#.

First, let's evaluate the magnitudes of both vectors:

#Rightarrow |vec(v)| = sqrt((4.0 times 10^(6))^(2) + (6.0 times 10^(6))^(2))#

#Rightarrow |vec(v)| = sqrt(52,000,000,000,000)#

#therefore |vec(v)| = 7,211,102.551#

#and#

#Rightarrow |vec(B)| = sqrt(0.030^(2) + (- 0.15)^(2))#

#Rightarrow |vec(B)| = sqrt(0.0234)#

#therefore |vec(B)| = 0.1529705854#

Then, let's find the angle between these two vectors:

#Rightarrow cos(theta) = frac(vec(v) cdot vec(B))(|vec(v)| cdot |vec(B)|)#

#Rightarrow cos(theta) = frac((4.0 times 10^(6) times 0.030) + (6.0 times 10^(6) times (- 0.15)))(7,211,102.551 cdot 0.1529705854)#

#Rightarrow cos(theta) = frac(120,000 + (- 900,000))(1,103,086.5786)#

#Rightarrow cos(theta) = frac(- 780,000)(1,103,086.5786)#

#Rightarrow cos(theta) = - 0.7071067812#

#Rightarrow theta = 2.35619449#

#therefore theta = 135^(circ)#

Now, let's use the formula #F = q v B sin(theta)#; where #q# is the charge of the proton, #v# is it's velocity, and #theta# is the angle between the velocity and magnetic field vectors:

#Rightarrow F = (1.6 times 10^(- 19)) cdot (7,211,102.551) cdot (0.1529705854) cdot sin(135^(circ))#

#Rightarrow F = 1.6 times 10^(- 19) cdot 1,103,086.5786 cdot 0.70710678118#

#Rightarrow F = 1.6 times 10^(- 19) cdot 780,000#

#therefore F = 1.248 times 10^(- 13)#

Therefore, the force on the proton is around #1.25 times 10^(- 13)# #"N"#.

#"b)"#

For this question, let's use the equation #frac(F)(L) = frac(mu_(0) I^(2))(2 pi d)#; where #frac(F)(L)# is the force per unit length, #mu_(0)# is the magnetic constant #(4 pi times 10^(- 7))#, #I# is the current, and #d# is the distance between the two wires:

#Rightarrow frac(F)(L) = frac(4 cdot pi times 10^(- 7) cdot 17^(2))(2 cdot pi cdot 0.002)#

#Rightarrow frac(F)(L) = frac(2 times 10^(- 7) cdot 289)(0.002)#

#Rightarrow frac(F)(L) = 0.0001 cdot 289#

#therefore frac(F)(L) = 0.0289#

Therefore, the force per unit length between the wires is #2.89 times 10^(- 2)# #"N/m"#.

Sep 9, 2017

a: #F=0.42*10^6veck#
b: #F/l=0.0289#

Explanation:

Let us first solve for (a)

we should know some vector properties

for this problem ,

#veci xxvecj=veck#

#vecjxxveci=-veck#
#vecixxveci=0#
#vecjxxvecj=0#

Okay Let's move on to the problem
Here it is given a magnetic field in x and y direction and also the velocity in in x and y direction

we use the relation of lorentz force
which is
#F=qE+q*(vecvxxvecB)#

since there is no electric field #E=0#
Hence #F=q*(vecvxxvec B)#

Where ,
q is the charge of the particle,
v is the velocity of the particle and
B is the magnetic field

Now just vector multiplication of #color(red)vecv# and #color(red)vecB#

#color(red)(vecvxxvecB)=(4.0*10^6veci+6.0*10^6vecj)xx(0.030veci+0.15vecj)#

Expanding the equation we get # color (red)(vec vxxvecB)=(4.0*10^6*0.030)(vecixxvecj)+(6.0*10^6*0.030)(vecjxxveci)+(4.0*10^6*0.15)(vecixxvecj)+(6.0*10^6*0.15)(vecjxxvecj)#

#color(red)(vecvxxvecB)=0+(6.0*10^6*0.030)(-veck)+(4.0*10^6*0.15)(veck)+0#

On summing up we get #color(red)(vecvxxvecB)=(0.6*10^6-0.18*10^6)veck#

Hence the magnetic force on the proton is given by

#F=1*(0.42*10^6)veck#
#color(blue)(q=1 )# Since the charge of the proton is one

Hence the answer will be #color(green)(F=0.42*10^6veck#

Okay let's move on to (b)

Force between  two parallel current carrying wires is given by

#F=(muI_1I_2l)/(2pid)#
where

#mu# is the permeability of free space which is equal to #4pi*10^-7#newton/ #ampere^2#

#I_1# is the current in first wire
#I_2# is the current in the second wire
#l # is the total length of the wires
#d# is the distance between the two wires

Then the force per unit length is given by
#F/l=(mu*I_1*I_2)/(2pi*d)#

substitute the values to get the answer

#I_1=I_2=17A#
#d=2*10^-3m#
#color(red)(F/l)=(4pi*10^-7*17*17)/(2*pi*2*10^-3#

#color(red)(F/l)=17^2*10^-4#
#color(red)(F/l)=0.0289# N/m