Numericals Physics

Question no.1 Suppose that an electron is accelerated from rest through a voltage difference of V=103 volts and then passes into a region containing a uniform magnetic field of magnitude B=1.2T.the electron subsequently executes a closed circular orbit in the plane perpendicular to the field. 1. What is the radius of the orbit? 2. What is the angular frequency of gyration of electron?

Solution: Given data:

∆V=103 volts B=1.2 T Ѳ=90° TO FIND:

RADIUS OF ORBIT=R=? ANGULAR FREQUENCY OF GYRATION=? FORMULA USED:

○ R=mveB ○ v=eBRm CALCULATIONS:

a) WE KNOW THAT qV=12mv2 So, velocity is equal to v=2qVm SO PUTTING VALUES IN THIS FORMULA v=2×103V×1.67×10-19 C9.1×10-31 kg

v=1.9×107m/sec

We know that R=mveB

R=9.1×10-31kg×1.9×107m/sec1.67×10-19C×1.2T= 8.69 ×10-5 m b) WE KNOW THAT v=eBRm SO PUTTING VALUES IN THIS FORMULA v=1.67×10-19C×1.2T×8.69×10-5m9.1×1031kg=1.91×107msec AS WE KNOW THAT v=Rω THEN ω=VR SO PUTTING VALUES IN THIS FORMULA ω=1.91×107msec8.69 ×10-5 m=2.11×1011 rad/sec Angular frequency (or angular speed) is the magnitude of the vector quantity angular velocity.

Question no.2 A plane circular loop of conducting wire of radius r =10 cm which possesses N=15 turns is placed in a uniform magnetic field. The direction of magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field strength is increased from B =1 T to B=5T in a time interval of ∆t= 10 sec. 1) What is the Emf generated around the loop? 2) If the electrical resistance of loop is R =15Ω, what current flows around the loop as the magnetic field is increased?

Solution: Given data: Radius = r=10 cm No. of turns= N=15 turns Angle=Ѳ=30° ∆B= 5 T – 1 T =4 T ∆t=10 second R =15Ω

To Find: ➢ Emf generated in loop =? ➢ Current flowing through loop =?

Formula used: •

ε=-∆Φ∆t=B.A∆t=BACOSѲ∆t

• V=IR Calculations: a) As we know that ε=-N∆Φ∆t=-N∆B.A∆t=-NBACOSѲ∆t

Putting the values in this formula ε=-15×4 T×π×0.12 10cos30 ε=-0.1639V b) As we know that

V=IR Since voltage is the induced Emf, so ε= I=εR Putting the values in this formula I=0.1639 V15 Ω I=0.0109 ampere

IR

Question no.3 Suppose that a conducting rod having length =0.2 m moving with speed v=0.1 m/s, under the magnetic field strength of B=1.0 T and the resistance of the circuit is R=0.020Ω. 1. What is the Emf generated around the circuit? 2. What current flows around the circuit? 3. What is the magnitude and direction of force acting on moving rod due to the fact that a current is flowing along it? 4. What is the rate at which work must be performed on the rod in order to keep it moving at constant velocity against this force? 5. What is the rate at which electrical energy is generated? 6. What is the rate at which energy is converted into heat due to the resistivity of the circuit?

Solution: Given data: =0.2 m v=0.1 m/s B=1.0 T R=0.020Ω Formulae used:



○ ○ ○

ε=-vB V=IR P=F.v

○

H=I2Rt

CALCULATIONS: a. AS WE KNOW THAT ε=-vB

sinѲ

BY TAKING THE ANGLE AS 90° ε=-vB



By putting the values in formula ε=-0.1m/s×1T×0.2m ε=-0.02 V Negative sign shows that induced Emf is always so as to oppose the change producing it. b. Current flowing through the conductor: I=VR Since voltage is the induced Emf, So I=εR Putting the values I=0.02 V0.020 Ω=1 amp c. FORCE ON CONDUCTOR F=ILBsinѲ F=1amp×0.2m×1T×sin90 F=0.2 N THE DIRECTION OF FORCE IS ALWAYS OPPOSITE TO THAT OF VELOCITY IN ORDER TO OBEY LENZ’S LAW AND TO CONSERVE LAW OF CONSERVATION OF ENERGY. d. RATE OF WORK DONE IS Wt=F.v Wt=0.2N×0.1m/sec Wt=0.02 J/S

e. RATE OF GENERATION OF ELECTRICAL ENERGY P=VI P=0.02V×1amp

p=0.02 J/S

AS RATE OF GENERATION OF ELECTRICAL ENERGY IS SAME AS THAT OF RATE OF CONSUMPTION OF ELECTRICAL ENERGY. f.

RATE OF HEAT GENERATION

Ht=I2Rtt Ht=I2R Ht=12×0.020Ω Ht=0.020 J/s