2(a) A positive charge q = 3.2 X 10^-19 C moves with a velocity v (2i+ 3j - k) m

Question

2(a) A positive charge q = 3.2 X 10^-19 C moves with a velocity v (2i+ 3j - k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving charge (In unit-vector notation) If B = (2 i + 4J + k) T, and E (4i -j - 2k) V/m. (b) What angle does the force vector make relative to the positive x axis? 2(b) Over a certain region of space. the electric potential is given by V = x5 - 3x^2y + 2yz^2. Find the expressions for the x, y, and: components of the electric field over this region. What Is the magnitude of the field at the point P, which has coordinates (In meter) (1,0,-2)?

Explanation / Answer

Q2.a

given that, charge=q=3.2*10^(-19) C

veloicty vector=2 i + 3j -k

B=2i + 4j +k

E=4i-j-2k

then total force=q*E+q*(cross product of v and B)

=3.2*10^(-19)*(4i-j-2k)+3.2*10^(-19)*(cross product of (2 i + 3j -k) and (2i + 4j +k))

=3.2*10^(-19)*(4i-j-2k)+3.2*10^(-19)*(7i-4j+2k)

=3.2*10^(-19)*(11 i-5 j)

hence total force is given by (3.52*10^(-18) i - 1.6*10^(-18) j) N

unit vector along the force vector=(11 i - 5 j)/sqrt(11^2+5^2)

=(0.91037 i - 0.4138 j)

let angle with +ve x axis is theta

then cos(theta)=dot product of (0.91037 i -0.4138 j) and i

(as both are unit vectors, their magnitudes are 1.)

hence cos(theta)=0.91037

==>theta=24.443 degree

Q2.b)

as we know that E=-grad (V)

=-(dv/dx) i -(dv/dy) j -(dv/dz) k

(all derivatives are partial derivatives)

given that v=5x-3*x^2*y+2*y*z^2

==>dv/dx=5-6*x*y

dv/dy=-3*x^2+2*z^2

dv/dz=4*y*z

then electric field=(6*x*y-5) i +(3*x^2-2*z^2) j - 4*y*z k

at x=1,y=0 and z=-2

electric field

=(6*1*0-5) i + (3*1^2-2*(-2)^2) j - 4*0*(-2) k

=-5 i -5 j

magnitude of electric field=sqrt(5^2+5^2)=7.071 V/m

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.