2. A small solid gold sphere with a diameter of 5.76mm has a kinetic energy of 3

Question

2. A small solid gold sphere with a diameter of 5.76mm has a kinetic energy of 300.0J and a charge of +4.44mC. The sphere is placed in a vacuum near the surface of the Earth where there is both a uniform magnetic field and a uniform electric field. The sphere is moving in a horizontal circle with a radius of 2.00m. The circular motion of the sphere is constant: it doesn't move up or down and the radius of the circle does not change. a) What is the magnitude and direction of the electric field needed to achieve this motion? b) What is the magnitude and direction of the magnetic field needed to achieve this motion?

Explanation / Answer

the electric force will be balanced by the weight of the sphere

and magnetic force supports the required centripetal force

Hence the Electric field and magnetic field are perpendicular to each other

Electric force = weight

q*E = m*g

4.44*10^-3*E = m*g

mass m = density *Volume = 19.6*1000*(4/3)*3.142*(5.76*10^-3/2)^3 = 1.96*10^-3 Kg

g = 9.8 m/s^2

then

q*E = m*g

E = m*g/q = 1.96*10^-3*9.8/(4.44*10^-3) = 4.32 N/C

direction is vertically upwards

b) Magnetic force = centripetal force

q*v*B = m*v^2/r

q*B = =m*v/r

4.44*10^-3*B = 1.96*10^-3*v/2

Given that Kinetic energy KE = 0.5*m*v^2 = 300

0.5*1.96*10^-3*v^2 = 300

v = 553.3 m/sec

then

4.44*10^-3*B = 1.96*10^-3*v/2

4.44*10^-3*B = 1.96*10^-3*553.3/2

B = 122.12 T direction is perpendicular to the plane of the horizontal circle

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