Question

the position 5. Consider a electron as Hydrogen atom with radius R = 5.29x10"m. current loop. If Treat the on orbiting this electron-proton system is placed in a magnetic field of 0.400T which is perpendicular to the magnetic moment of the loop, what is the torque ? (me = 9.11*10-31 kg , 1968- 1.60*10'SC)

with required formula

Answer 1

Torque   .........................(1)

T=μXB

where $\mu$ is magnetic moment of current loop (electron-proton system of hydrogen atom ) and B is magnetic field induction .

Since it is given that magnetic moment is perpendicular to magnetic field induction ,

torque $\tau$ = ( $\mu$ B )......................................(2)

magnetic moment of current loop, $\mu$ = I $\times$ A ...................................(3)

where I is current in the loop and A is area of loop.

Equivalent current I passing in the loop formed by electron-proton system,

I = (q v ) /(2$\pi$r) ....................................(4)

where q is charge of electron , v is speed of electron and r is radius of electron orbit .

speed of electron is obtained from the follwing equation

....................................(5)

,2 e K 11 2 7

where LHS of above equation is Coulomb force and RHS of above equation is Centripetal force

K = 9 $\times$ 10⁹ N m² C^-2 is coulomb constant, e = 1.602 $\times$ 10^-19 C is charge on electron and                          m = 9.1 $\times$ 10^-31 kg is mass of electron

By substituting values in eqn.(5), we get speed v of electron in orbit as , v = 2.19 $\times$ 10⁶ m/s

Current is calculated from eqn.(4) as , I = 1.055 $\times$ 10^-3 A

Magnetic moment = $\mu$ = I $\times$ A = 1.056 $\times$ 10^-3$\times$$\pi$$\times$ 5.29 $\times$ 5.29 $\times$ 10^-22 = 9.284 $\times$ 10^-24 A/m²

Torque = $\mu$ B = 9.284 $\times$ 10^-24$\times$ 0.4 = 3.714 $\times$ 10^-24 N m