At one instant an electron(charge = -1.6 x 10-19 C)is moving in the xy plane ,...

At one instant an electron(charge = -1.6 x 10^-19 C)is moving in the xy plane , the components of its velocity beingv_x = 3.0 x 10⁵ m/s and v_y=4.0x10⁵m/s. A magnetic field of 0.8T is in the positivez direction. At that instant the magnitude of the magnetic force onthe electron is ....

(a) 0

(b) 2.9 x 10^-14 N

(d) 4.8 x 10^-14 N

(e) 6.4x10^-14 N

Solutions

Expert Solution

v =v_xi+v_yj =3.0 x 10⁵i +4.0x10⁵j m/s

B = 0.8 k

q = -1.6 x 10^-19 C

F = q(v X B)= -1.6 x10^-19(3.0 x10⁵i + 4.0x10⁵j) X 0.8 k
= -1.6 x 10^-19 *3.0 x 10⁵ *0.8(-j) + -1.6 x10^-19 *4.0x10⁵*0.8(-i)
= 3.84 x 10^-14j + 5.12 x10^-14i [ Sincei X k = -j andj X k = -i]

∴ |F| = √[( 3.84 x 10^-14)² +(5.12 x10^-14)²] = 6.4 x10^-14 N

genius_generous answered 2 years ago

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