At one instant, m/s is the velocity of a proton in a uniform magnetic field mT. At that instant, what are (a) the magnetic force acting on the proton, in unit-vector notation and (b) the angle between and .

Complete question is;

At one instant, v = (−2.00 i^ +4 .00 j^ − 6.00 k^)m/s is the velocity of a proton in a uniform magnetic field B = (2.00 i^ − 4.00 j^ +8 .00 k^) mT. At that instant, what are;

(a) the magnetic force acting on the proton, in unit-vector notation

(b) the angle between the velocity and magnetic field

A) F = [1.28 × 10^(-21)] i^ + [6.4 × 10^(-22)] j^

B) θ = 151.02°

Explanation:

A) Since the unit of magnetic field is given in mT, let's convert it to T.

Thus;

B = (0.002 i^ − 0.004 j^ + 0.008 k^) T

Formula for the magnetic force is;

F = qVB

Where q = 1.6 × 10^(-19) C

Since V and B are in vector notation, the way to multiply it is;

(v_y•B_z - v_z•B_y)i^ + (v_z•B_x - v_x•B_z) j^ + (v_x•B_y - v_y•B_x) k^

Thus;

F = 1.6 × 10^(-19)((v_y•B_z - v_z•B_y)i^ + (v_z•B_x - v_x•B_z) j^ + (v_x•B_y - v_y•B_x) k^)

Plugging in the relevant values;

F = 1.6 × 10^(-19)[{(4 × 0.008) - (-6 × -0.004)} i^ + {(-6 × 0.002) - (-2 × 0.008)} j^ + {(−2 × -0.004) - (4 × 0.002)} k^]

Multiplying out gives;

F = [1.28 × 10^(-21)] i^ + [6.4 × 10^(-22)] j^

B) Angle between the 2 vectors gotten for the velocity and magnetic field is given by the formula;

Cos θ = (a•b)/(|a| × |b|)

(a•b) = (−2.00 i^ + 4.00 j^ − 6.00 k^) • (0.002 i^ − 0.004 j^ + 0.008 k^)

>> (2 × 0.002) + (4 × -0.004) + (-6 × 0.008) = -0.06

|a| = √((-2)² + (4)² + (-6)²)

|a| = √56

|b| = √((0.002)² + (-0.004)² + (0.008)²)

|b| = 0.009165

Thus;

Cos θ = -0.06/(0.009165 × √56)

Cos θ = -0.8748

θ = cos^(-1) -0.8748

θ = 151.02°

true is the correct answer

hope i