Question

A particle with charge q= 7.80?C is moving with velocity v= -3.8 E 3 j [m/s]. The magnetic force is measured to be F= (+7.60 E -3 i - 5.20 E -3 k) [N].

1. Calculate the components of the magnetic field
2. Can there be components of the magnetic field that are not determined by the Force?
3. Calculate the vector dot product F*B, what is the angle between F and B?

Answer 1

We know that magnetic force on a moving charge in magnetic field is given by:

F = q*VxB

V = (-3800 j) m/s

Assuming, B = (Bx i + By j + Bz k) T

q = charge on particle = 7.80*10^-6 C

Given that, F = (7.60*10^-3 i - 5.20*10^-3 k) N

Now

F = q*VxB = 7.80*10^-6*(-3800 j)x((Bx i + By j + Bz k))

F = -3800*7.80*10^-6*Bx (jxi) - 3800*7.80*10^-6*By (jxj) - 3800*7.80*10^-6*Bz (jxk)

F = -29.64*10^-3*Bx (-k) - 0 - 29.64*10^-3*Bz (i)

F = -29.64*10^-3*Bz i + 29.64*10^-3*Bx k

Now comparing both force values:

(7.60*10^-3 i - 5.20*10^-3 k) N = (-29.64*10^-3*Bz i + 29.64*10^-3*Bx k) N

-29.64*10^-3*Bz = 7.60*10^-3

Bz = 7.60*10^-3/(-29.64*10^-3) = -0.256 T

29.64*10^-3*Bx = 5.20*10^-3

Bx = 5.20*10^-3/(29.64*10^-3) = 0.175 T

Bx = x component of magnetic field = 0.175 T

Bz = z component of magnetic field = -0.256 T

Part B.

Since there is no force on particle in y-direction, So magnetic field cannot be determined in y-direction

Part C.

Scalar product between B and F will be:

B = (0.175 i - 0.256 k) T

F = (7.60*10^-3 i - 5.20*10^-3 k) N

B.F = (0.175 i - 0.256 k).(7.60*10^-3 i - 5.20*10^-3 k)

In scalar product, i.i = k.k = 1

B.F = 0.175*7.60*10^-3 + 0.256*5.20*10^-3

B.F = (2.66*10^-3) N-T = Dot product of F and B

We know that in scalar product:

B.F = |B|*|F|*cos theta

theta = arccos (B.F/(|B|*|F|))

|B| = sqrt (0.175^2 + (-0.256)^2) = 0.310

|F| = 10^-3*sqrt (7.60^2 + (-5.20)^2) = 9.209*10^-3

theta = arccos (2.66*10^-3/(0.310*9.209*10^-3))

theta = direction between B and F = 21.288 deg = 21.3 deg

Let me know if you've any query.