What is the kinetic energy of an electron that posses undedicated through perpen

Question

What is the kinetic energy of an electron that posses undedicated through perpendicular electric and magnetic fields if E = 44 kV/m and B = 0.8mI? Two charged particles enter a magnetic field, directed out of the page, from the top and take the paths shown in the figure. Which statement is correct? Two long parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the magnetic field at a point in the plane of the wires an d10 cm from each wire is 4.0 mu T. What is the larger of the two currents ? At a location near the equator, the earth's magnetic field is horizontal, i.e., parallel to the earth's surface, and points north. An electron is moving upward from the ground. What is the direction of the force that acts on the electron?

Explanation / Answer

1) For electron to go undeviated

eE = e(v X B)

v = E/B = 44000/0.8*10^(-3) = 5.5*10^(7) m/sec

KE of electron =0.5mv^2= 0.5*9.1*10^(-31) * (5.5 * 10^7)^2 = 1.4*10^(-15) J

Option C is correct

2)according to maxwell right hand rule

F = e(V X B)

This means if charge is positive , it will deviate right of its original direction ,otherwise left

So 1 is positive & 2 is negative

Option A is correct (ans)

3)The magnetic field due to both wires at 10 cm from both wire will be in opposite direction

So B = munot (I1 - I2) /2pid = 2*10^(-7) * (3I - I) /d = 4*10^(-7) * I/0.1 = 4*10^(-6) *I = 4 * 10^(-6)T

So I =1A

So larger current =3I = 3A (ans)

Option B is correct

4) According to maxwells right hand rule ,F will be acting towards west

So option D is correct (ans)

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