wayground logo

Free Printable Worksheets

NEW

Font size

S
M
L
XL
Worksheets

PROBLEM FORMULA TEST - VOL 1

Total questions: 52

Worksheet time: 39mins

Name
Class
Date
1.

Calculate the number of electrons in one
coulomb of negative charge.

a)

Q =NE

b)

N=QE

c)

n = qe\frac{\text{q}}{\text{e}}  

d)

e =  nq\frac{n}{q}  

2.

Consider two point charges q1 and q2. They are separated by a distance of 1m. Calculate the force experienced by the two charges

a)

q1q2R\frac{q_1q_2}{R} k = f

b)

q1q2R2\frac{q_1q_2}{R^2} k = f

c)

q1q2R3\frac{q_1q_2}{R^3} k = f

d)

\frac{q_1q_2}{R^3} = f

3.

Consider four equal charges q1, q2, q3

and q4 = q = +1 μC located at four different points

on a circle of radius 1m. Calculate the total force acting on the charge q1

due to all the other charges.

a)

F = F1F_1 + F2F_2 + F3F_3 + F4F_4

b)

F1tot F_1^{tot}\ = F_1 + F_2 + F_3

c)

F_1^{tot}\ = F11F_{11} + F21F_{21} + F31F_{31} + F41F_{41}

d)

F_1^{tot}\ = F12F_{12} + F13F_{13} + F14F_{14}

4.

Calculate the electric field at points P

a)

EPE_P = K QR\frac{Q}{R}

b)

E_P = K Q2R\frac{Q^2}{R}

c)

E_P = K Q2R2\frac{Q^2}{R^2}

d)

E_P = K QR2\frac{Q}{R^2}

5.

A block of mass m carrying a positive charge

q is placed on an insulated friction less

inclined plane as shown in the figure. A

uniform electric field E is applied parallel

to the inclined surface such that the block

is at rest. Calculate the magnitude of the

electric field E.

a)

E = Fq\frac{F}{q}

b)

E = Fq2\frac{F}{q^2}

c)

E = F2q2\frac{F^2}{q^2}

d)

E = k \frac{F^2}{q^2}

6.

Calculate the electric dipole moment for the following charge configurations.

a)

P = i=1nqiri\sum_{i=1}^nq_ir_i

b)

p = qr

c)

p = q2q^2 r

d)

p = q r2r^2

7.

a)

 τ\tau  = pE  cosθ\cos\theta  

b)

 \tau  = pE  sinθ\sin\theta  

c)

 \tau  = 2pE  \cos\theta  

d)

 \tau  = 2pE  sinθ\sin\theta  

8.

The following figure represents the electric potential as a function of x – coordinate. Plot the corresponding electric field as a function of x.

a)

E = dxdv\frac{\text{d}x}{\text{d}v}

b)

E = - \frac{\text{d}x}{\text{d}v}

c)

E = - dvdx\frac{\text{d}v}{\text{d}x}

d)

E = \frac{\text{d}v}{\text{d}x}

9.

Four charges are arranged at the corners of the square PQRS of side a as shown in the figure.(a) Find the work required to assemble these charges in the given configuration. (b) Suppose a charge q′ is brought to the centre of the square, by keeping the four charges fixed at the corners, how much extra work is required for this?

a)

W = q F

b)

W = F d

c)

W = q V

d)

W = V d

10.
a)

U = - pE cosθ\cos\theta

b)

U = - qE \cos\theta

c)

U = - pE sinθ\sin\theta

d)

U = - qE sinθ\sin\theta

11.

a)

ϕE\phi_E = E . F cosθ\cos\theta

b)

\phi_E = E . A sinθ\sin\theta

c)

\phi_E = E . A \cos\theta

d)

\phi_E = E . F sinθ\sin\theta

12.

A small ball of conducting material having a charge +q and mass m is thrown upward at an angle θ to horizontal surface with an initial speed vo as shown in the figure. There exists an uniform electric field E downward along with the gravitational field g. Calculate the acceleration in the motion of this charged ball.

a)

a = qim\frac{qi}{m}

b)

a = qem\frac{qe}{m}

c)

a = qEm\frac{qE}{m}

d)

a = qIm\frac{qI}{m}

13.
a)

C = ϵ0 Ad\frac{\epsilon_{0\ A}}{d} , C=QV

b)

C = ϵ0 Ed\frac{\epsilon_0\ E}{d} , Q=C V

c)

C = \frac{\epsilon_{0\ A}}{d} , Q=C V

d)

C = \frac{\epsilon_{0\ A}}{d} , V=Q C

14.

a)

C = ϵmAd\frac{\epsilon_mA}{d}

b)

C = ϵ0Ad\frac{\epsilon_0A}{d}

c)

C = ϵr Ad\frac{\epsilon_{r\ }A}{d}

15.

a)

C = C1 + C2

b)

C = C1 . C2

c)

C= 1C1\frac{1}{C_1} + 1C2\frac{1}{C_2}

d)

C= \frac{1}{C_1} . \frac{1}{C_2}

16.

Compute the current in the wire if a charge of 120 C is flowing through a copper wire in 1 minute.

a)

Q = I t

b)

I = Q t

c)

Q = It\frac{I}{t}

d)

t = QI\frac{Q}{I}

17.

If an electric field of magnitude 570 N C–1, is applied in the copper wire, find the acceleration experienced by the electron.

a)

a = Fm\frac{F}{m}

b)

a = eEm\frac{eE}{m}

c)

a = Fq\frac{F}{q}

d)

a = qF\frac{q}{F}

18.
a)

VdV_d = IeE\frac{I}{eE}

b)

V_d = InE\frac{I}{nE}

c)

V_d = IneA\frac{I}{neA}

d)

V_d = IneE\frac{I}{neE}

19.

The resistance of a wire is 20 Ω. What will be new resistance, if it is stretched uniformly 8 times its original length?

a)

ρ\rho = R lA\frac{l}{A}

b)

\rho = R Al\frac{A}{l}

c)

RR = ρ \frac{l}{A}

d)

R = ρ Al\frac{A}{l}

20.

Calculate the equivalent resistance for the circuit which is connected to 24 V battery and also find the potential difference across each resistors in the circuit.

a)

V = IR, R = R1 + R2R_1\ +\ R_2

b)

R = IV, R = R_1\ +\ R_2

c)

V = IR, R = 1R1 + 1R2\frac{1}{R_1}\ +\ \frac{1}{R_2}

d)

R = IV, R = \frac{1}{R_1}\ +\ \frac{1}{R_2}

21.
a)

RT=R0(1+α (TT0))R_T=R_0(1+\alpha\ (T-T_0))

b)

R0=RT(1+α (TT0))R_0=R_T(1+\alpha\ (T-T_0))

c)

RT=R0(1+(αTT0))R_T=R_0(1+(\alpha T-T_0))

d)

RT=R0(α+α(TT0))R_T=R_0(\alpha+\alpha(T-T_0))

22.

a)

α =1R0 ΔRΔT \alpha\ =\frac{1}{R_0}\ \frac{\Delta R}{\Delta T}\

b)

α =1T ΔTΔR0\alpha\ =\frac{1}{T}\ \frac{\Delta T}{\Delta R_0}

c)

α =1T0 ΔRΔT\alpha\ =\frac{1}{T_0}\ \frac{\Delta R}{\Delta T}

d)

α =1RT ΔRΔT \alpha\ =\frac{1}{R_T}\ \frac{\Delta R}{\Delta T}\

23.

A battery of voltage V is connected to 30 W bulb and 60 W bulb as shown in the figure. (a) Identify brightest bulb

(b) which bulb has greater resistance? (c) Suppose the two bulbs are connected in series, which bulb will glow brighter?

a)

V =I R

b)

V = I P

c)

P = I R

d)

P = I V

24.

Two electric bulbs marked 20 W – 220 V and 100 W – 220 V are connected in series to 440 V supply. Which bulb will

get fused?

a)

R =V2PR\ =\frac{V^2}{P}

b)

V =I2PV\ =\frac{I^2}{P}

c)

P =R2VP\ =\frac{R^2}{V}

d)

P =I2VP\ =\frac{I^2}{V}

25.

A battery has an emf of 12 V and connected to a resistor of 3 Ω. The current in the circuit is 3.93 A. Calculate power delivered by the battery and power delivered to the resistor

a)

r =ϵ VV Rr\ =\left|\frac{\epsilon\ -\ V}{V}\right|\ R

b)

r =V ϵV Rr\ =\left|\frac{V\ -\ \epsilon}{V}\right|\ R

c)

r = V ϵϵ Rr\ =\ \left|\frac{V\ -\ \epsilon}{\epsilon}\right|\ R

d)

r = ϵ Vϵ Rr\ =\ \left|\frac{\epsilon\ -\ V}{\epsilon}\right|\ R

26.

Let the magnetic moment of a bar magnet be p whose magnetic length is d = 2l and pole strength is qm. Compute the magnetic moment of the bar magnet when it is cut into two pieces

a)

P = 2ql

b)

P= 2qE

c)

P = 2ld

d)

P= 2Vl

27.

A short bar magnet has a magnetic moment of 0.5 J T –1. Calculate magnitude and direction of the magnetic field

produced by the bar magnet which is kept at a distance of 0.1 m from the centre of the bar magnet along (a) axial line of the bar magnet and (b) normal bisector of the bar magnet.

a)

Baxial = μ04π 2Pmr3B_{axial}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{2P_m}{r^3}\right| , Baxial = μ04π Pmr2B_{axial}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{P_m}{r^2}\right|

b)

Baxial = μ04π 2Pmr2B_{axial}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{2P_m}{r^2}\right| Bequ = μ04π 2Pmr3B_{equ}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{2P_m}{r^3}\right|

c)

Baxial = μ04π 2Pmr3B_{axial}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{2P_m}{r^3}\right| Bequ = μ04π Pmr3B_{equ}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{P_m}{r^3}\right|

d)

Baxial = μ04π 2PmrB_{axial}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{2P_m}{r^{ }}\right| Bequ = μ04π PmrB_{equ}\ =\ \frac{\mu_0}{4\pi}\ \left|\frac{P_m}{r}\right|

28.

Consider a magnetic dipole which on switching ON external magnetic field orient only in two possible ways

i.e., one along the direction of the magnetic field (parallel to the field) and another anti-parallel to magnetic field. Compute the energy for the possible orientation.

a)

Uparallel = Uminimum = Pm BcosθU_{parallel}\ =\ U_{\min imum}\ =\ -\ P_m\ B\cos\theta

b)

Uparallel = Umaximum = Pm BcosθU_{parallel}\ =\ U_{\max imum}\ =\ -\ P_m\ B\cos\theta

c)

Uparallel = Umaximum = Pm EcosθU_{parallel}\ =\ U_{\max imum}\ =\ \ P_m\ E\cos\theta

d)

Uparallel = Uminimum = Pm EcosθU_{parallel}\ =\ U_{\min imum}\ =\ -\ P_m\ E\cos\theta

29.
a)

χm,X = MH\chi_{m,X}\ =\ \frac{\left|\overrightarrow{M}\right|}{\left|\overrightarrow{H}\right|}

b)

χm,X = HM\chi_{m,X}\ =\ \frac{\left|\overrightarrow{H}\right|}{\left|\overrightarrow{M}\right|}

c)

χm,X = BH\chi_{m,X}\ =\ \frac{\left|\overrightarrow{B}\right|}{\left|\overrightarrow{H}\right|}

d)

χm,X = HB\chi_{m,X}\ =\ \frac{\left|\overrightarrow{H}\right|}{\left|\overrightarrow{B}\right|}

30.

a)

Using right hand rule, current flows upwards.

b)

Using right hand rule, current flows downwards.

c)

Using Fleming's right hand rule, current flows upwards.

d)

Using Fleming's right hand rule, current flows downwards.

31.

a)

B = 0

b)

B = 1

c)

B = Maximum

d)

B = Minimum

32.

a)

I = 2RBHμoN tanθI\ =\ \frac{2RB_H}{\mu_oN}\ \tan\theta

b)

I = 2RBHμoN cosθI\ =\ \frac{2RB_H}{\mu_oN}\ \cos\theta

c)

I = 2RμoN tanθI\ =\ \frac{2R}{\mu_oN}\ \tan\theta

d)

I = BHμoN tanθI\ =\ \frac{B_H}{\mu_oN}\ \tan\theta

33.

Compute the magnitude of the magnetic field of a long, straight wire carrying a current of 1 A at distance of 1m

from it. Compare it with Earth’s magnetic field.

a)

Bstraight wire = μo I2πrB_{straight\ wire}\ =\ \frac{\mu_o\ I}{2\pi r}

b)

Bstraight wire = I2πrB_{straight\ wire}\ =\ \frac{\ I}{2\pi r}

c)

Bstraight wire = μo I2rB_{straight\ wire}\ =\ \frac{\mu_o\ I}{2r}

d)

Bstraight wire = μo I2πB_{straight\ wire}\ =\ \frac{\mu_o\ I}{2\pi}

34.

Calculate the magnetic field inside a solenoid, when

a)

B = μ0NlAB\ =\ \frac{\mu_0Nl}{A}

b)

B = μ0NlAB\ =\ \frac{\mu_0N}{lA}

c)

B = μ0NILB\ =\ \frac{\mu_0NI}{L}

d)

B = μ0NLIB\ =\ \frac{\mu_0NL}{I}

35.

Compute the work done and power delivered by the Lorentz force on the particle of charge q moving with velocity v . Calculate the angle between Lorentz force

and velocity of the charged particle and also interpret the result.

a)

f = q (v X B) ; W = f . dr ; dWdt = Pf\ =\ q\ \left(\overrightarrow{v}\ X\ \overrightarrow{\ B}\right)\ ;\ W\ =\ \int_{ }^{ }\overrightarrow{f}\ .\ \overrightarrow{dr\ }\ ;\ \frac{dW}{dt}\ =\ P

b)

f = q (E X B) ; W = f . dr ; dWdt = Pf\ =\ q\ \left(\overrightarrow{E}\ \ X\ \overrightarrow{\ B}\right)\ ;\ W\ =\ \int_{ }^{ }\overrightarrow{f}\ .\ \overrightarrow{dr\ }\ ;\ \frac{dW}{dt}\ =\ P

c)

f = E (v X B) ; W = E . dr ; dWdt = Pf\ =\ E\ \left(\overrightarrow{v}\ X\ \overrightarrow{\ B}\right)\ ;\ W\ =\ \int_{ }^{ }\overrightarrow{E}\ .\ \overrightarrow{dr\ }\ ;\ \frac{dW}{dt}\ =\ P

d)

f = v (q X B) ; W = f . dr ; dWdt = Pf\ =\ v\ \left(\overrightarrow{q}\ X\ \overrightarrow{\ B}\right)\ ;\ W\ =\ \int_{ }^{ }\overrightarrow{f}\ .\ \overrightarrow{dr\ }\ ;\ \frac{dW}{dt}\ =\ P

36.

An electron moving perpendicular to a uniform magnetic field 0.500 T undergoes circular motion of radius 2.50 mm. What is the speed of electron?

a)

v = q rBfv\ =\ \left|q\right|\ \frac{rB}{f}

b)

v = q rmBv\ =\ \left|q\right|\ \frac{rm}{B}

c)

v = q rBmv\ =\ \left|q\right|\ \frac{rB}{m}

d)

v = B rfqv\ =\ \left|B\right|\ \frac{rf}{q}

37.
a)

v = ERv\ =\ \frac{E}{R}

b)

v = EBv\ =\ \frac{E}{B}

c)

v = BEv\ =\ \frac{B}{E}

d)

v = REv\ =\ \frac{R}{E}

38.

a)

mg sinθ = IBl cosθmg\ \sin\theta\ =\ IBl\ \cos\theta

b)

mg cosθ = IBl sinθmg\ \cos\theta\ =\ IBl\ \sin\theta

c)

mg tanθ = IBmg\ \tan\theta\ =\ IB

d)

mg tanθ = IBl cosθmg\ \tan\theta\ =\ IBl\ \cos\theta

39.

a)

Is = NABK ; Vs = θVI_s\ =\ \frac{NAB}{K}\ ;\ V_s\ =\ \frac{\theta}{V}

b)

Is = NKBA ; Vs = θVI_s\ =\ \frac{NKB}{A}\ ;\ V_s\ =\ \frac{\theta}{V}

c)

Is = NABK ; Vs = VθI_s\ =\ \frac{NAB}{K}\ ;\ V_s\ =\ \frac{V}{\theta}

d)

Is = NVBK ; Vs = θVI_s\ =\ \frac{NVB}{K}\ ;\ V_s\ =\ \frac{\theta}{V}

40.

a)

ϕB = Bl cosθ\phi_B\ =\ Bl\ \cos\theta

b)

ϕB = BA cosθ\phi_B\ =\ BA\ \cos\theta

c)

ϕB = Bl sinθ\phi_B\ =\ Bl\ \sin\theta

d)

ϕB = BA sinθ\phi_B\ =\ BA\ \ \sin\theta

41.

a)

ϕi = BA cosθ; ϵ = NdϕBdt\phi_{i\ }=\ BA\ \cos\theta;\ \epsilon\ =\ N\frac{d\phi_B}{dt}

b)

ϕi = BA sinθ; ϵ = NdϕBdt\phi_{i\ }=\ BA\ \sin\theta;\ \epsilon\ =\ N\frac{d\phi_B}{dt}

c)

ϕi = nA cosΘ; ϵ = NdϕBdt\phi_{i\ }=\ nA\ \cos\Theta;\ \epsilon\ =\ N\frac{d\phi_B}{dt}

d)

ϕi = BA cosΘ; ϵ = NdBdt\phi_{i\ }=\ BA\ \cos\Theta;\ \epsilon\ =\ N\frac{dB}{dt}

42.

a)

v2 = u2 + 2gl ; ϵ = BHlvv^2\ =\ u^2\ +\ 2gl\ ;\ \epsilon\ =\ B_Hlv

b)

v2 = u2 + 2gh ; ϵ = BHlev^2\ =\ u^2\ +\ 2gh\ ;\ \epsilon\ =\ B_Hle

c)

v2 = u2 + 2gh ; ϵ = BHlvv^2\ =\ u^2\ +\ 2gh\ ;\ \epsilon\ =\ B_Hlv

d)

v2 = u2 + 2g ; ϵ = BHlev^2\ =\ u^2\ +\ 2g\ ;\ \epsilon\ =\ B_Hle

43.

A solenoid of 500 turns is wound on an iron core of relative permeability 800. The length and radius of the solenoid are 40 cm and 3 cm respectively. Calculate the

average emf induced in the solenoid if the current in it changes from 0 to 3 A in 0.4 second.

a)

L = μn2Al ; ϵ = L didtL\ =\ \mu n^2Al\ ;\ \epsilon\ =\ -L\ \frac{di}{dt}

b)

L = μn2Al2 ; ϵ = L didtL\ =\ \mu n^2Al^2\ ;\ \epsilon\ =\ -L\ \frac{di}{dt}

c)

L = μnAl2 ; ϵ = L didtL\ =\ \mu nAl^2\ ;\ \epsilon\ =\ -L\ \frac{di}{dt}

d)

L = μnAl ; ϵ = L didtL\ =\ \mu nAl\ ;\ \epsilon\ =\ -L\ \frac{di}{dt}

44.

The self-inductance of an air-core solenoid is 4.8 mH. If its core is replaced by iron core, then its self-inductance becomes 1.8 H. Find out the relative permeability of

iron.

a)

Liron = μrLairL_{iron\ }=\ \mu_rL_{air}

b)

Lair = μrLironL_{air\ }=\ \mu_rL_{iron}

c)

1LironLair = μr\frac{1}{Liron}L_{air\ }=\ \mu_r

d)

Liron = 1μrLairL_{iron\ }=\ \frac{1}{\mu_r}L_{air}

45.
a)

ϵ = ϵ0 sinωt\epsilon\ =\ \epsilon_0\ \sin\omega t

b)

ϵ = ϵ0 cos ωt\epsilon\ =\ \epsilon_0\ \cos\ \omega t

c)

ϵ = E0 sinωt\epsilon\ =\ E_0\ \sin\omega t

d)

ϵ = E0 cosωt\epsilon\ =\ E_0\ \cos\omega t

46.

An ideal transformer has 460 and 40,000 turns in the primary and secondary coils respectively. Find the voltage developed per turn of the secondary if the transformer is connected to a 230 V AC mains. The

secondary is given to a load of resistance 104 Ω.

a)

Vs = Vp NsNpV_s\ =\ \frac{V_p\ N_s}{N_p}

b)

Vs = NsNpV_s\ =\ \frac{\ N_s}{N_p}

c)

Vs = Vp NpNsV_s\ =\ \frac{V_p\ N_p}{N_s}

d)

Vs = NpNsV_s\ =\ \frac{\ N_p}{N_s}

47.

An inverter is common electrical device which we use in our homes. When there is no power in our house, inverter gives AC power to run a few electronic appliances like fan or light. An inverter has inbuilt step-up transformer which converts 12 V AC to 240 V AC. The primary coil has 100 turns and the inverter delivers 50 mA to the external circuit. Find the number of turns

in the secondary and the primary current.

a)

VsVP = NsNp = IpIs\frac{V_s}{V_P}\ =\ \frac{N_s}{N_p}\ =\ \frac{I_p}{I_s}

b)

VsVP = NsNp = IsIp\frac{V_s}{V_P}\ =\ \frac{N_s}{N_p}\ =\ \frac{I_s}{I_p}

c)

VsVP = NpNs = IpIs\frac{V_s}{V_P}\ =\ \frac{N_p}{N_s}\ =\ \frac{I_p}{I_s}

d)

VpVs = NsNp = IpIs\frac{V_p}{V_s}\ =\ \frac{N_s}{N_p}\ =\ \frac{I_p}{I_s}

48.

Write down the equation for a sinusoidal voltage of 50 Hz and its peak value is 20 V. Draw the corresponding voltage versus time graph.

a)

v = Em sinωt ; ω = 2πfv\ =\ E_m\ \sin\omega t\ \ ;\ \omega\ =\ 2\pi f

b)

v = Vm cosωt ; ω = 2πfv\ =\ V_m\ \cos\omega t\ \ ;\ \omega\ =\ 2\pi f

c)

v = Vm sinωt ; ω = 2πfv\ =\ V_m\ \sin\omega t\ \ ;\ \omega\ =\ 2\pi f

d)

v = Em cosωt ; ω = 2πfv\ =\ E_m\ \cos\omega t\ \ ;\ \omega\ =\ 2\pi f

49.

Find the impedance of a series RLC circuit if the inductive reactance, capacitive reactance and resistance are 184 Ω, 144 Ω and 30 Ω respectively. Also calculate the phase angle between voltage and current.

a)

Z = R2+(XLXC)2 ; tanϕ = XLXCRZ\ =\ \sqrt{R^2+\left(X_L-X_C\right)^2}\ \ ;\ \tan\phi\ =\ \frac{X_L-X_C}{R}

b)

Z = R+(XLXC) ; tanϕ = XLXCRZ\ =\ \sqrt{R^{ }+\left(X_L-X_C\right)^{ }}\ \ ;\ \tan\phi\ =\ \frac{X_L-X_C}{R}

c)

Z = R2+XL2+XC2 ; tanϕ = XLXCRZ\ =\ \sqrt{R^2+X_L^2+X_C^2}\ \ ;\ \tan\phi\ =\ \frac{X_L-X_C}{R}

d)

Z = R2(XLXC)2 ; tanϕ = XLXCRZ\ =\ \sqrt{R^2-\left(X_L-X_C\right)^2}\ \ ;\ \tan\phi\ =\ \frac{X_L-X_C}{R}

50.

A series RLC circuit which resonates at 400 kHz has 80 μH inductor, 2000 pF capacitor and 50 Ω resistor. Calculate Q-factor of the circuit

a)

Q = 1LRCQ\ =\ \frac{1}{L}\sqrt{\frac{R}{C}}

b)

Q = 1RLCQ\ =\ \frac{1}{R}\sqrt{\frac{L}{C}}

c)

Q = 1CLRQ\ =\ \frac{1}{C}\sqrt{\frac{L}{R}}

d)

Q = 1RLCQ\ =\ \frac{1}{R\sqrt{\frac{L}{C}}}

51.

The relative magnetic permeability of the medium is 2.5 and the relative electrical permittivity of the medium is

2.25. Compute the refractive index of the medium.

a)

n = 1ϵmμmn\ =\ \frac{1}{\sqrt{\epsilon_m\mu_m}}

b)

n = ϵ0μ0n\ =\ \sqrt{\epsilon_0\mu_0}

c)

n = ϵrμrn\ =\ \sqrt{\epsilon_r\mu_r}

d)

n = 1ϵrμrn\ =\ \frac{1}{\sqrt{\epsilon_r\mu_r}}

52.

A magnetron in a microwave oven emits electromagnetic waves (em waves) with frequency f = 2450 MHz. What magnetic field strength is required for electrons to

move in circular paths with this frequency?.

a)

B = qωmeB\ =\ \frac{\left|q\right|\omega}{m_e}

b)

B = meqωB\ =\ \frac{m_e\left|q\right|}{\omega}

c)

B = meωq2B\ =\ \frac{m_e\omega}{\left|q^2\right|}

d)

B = meωqB\ =\ \frac{m_e\omega}{\left|q\right|}