Compare penetration and ionization abilities

Consider safety requirements for food

UV provides sterilization without harmful effects

Carrier wave fc modulated by signal fm

Produces frequencies fc+fm and fc-fm

Results in carrier and two sidebands

LOS requires direct path without obstacles

Space waves travel in straight lines

Best suited for LOS communication

Use Curie law: χT = constant

χ₁T₁ = χ₂T₂ where T in Kelvin

0.0075(200) = χ₂(100), so χ₂ = 0.015

Electromagnet cores need to be easily magnetized (high permeability)

Should lose magnetism easily when current stops (low retentivity)

Therefore need high permeability and low retentivity

F/l = (μ₀/2π)×(I₁I₂/r) where μ₀/2π = 10⁻⁷

Substitute I₁=1A, I₂=3A, r=1m

F/l = 10⁻⁷×(1×3)/1 = 6×10⁻⁷ N/m; Opposite currents repel

Cyclotron uses magnetic and electric fields

Both +ve and -ve charges experience required forces

Direction of rotation differs but acceleration possible for both

Use r = mv/qB where q/m is given

Substitute v = 10⁹ m/s, B = 10⁻⁴ T, q/m = 10¹¹ C/kg

r = (10⁹)/(10¹¹×10⁻⁴) = 100 m

B₁ = μ₀I₁/2r and B₂ = μ₀I₂/2r for each coil

Since coils are perpendicular, use Pythagoras B = √(B₁² + B₂²)

Calculate with given values to get 5×10⁻⁵ Wb/m²

Understand that carrier wave is modified according to message signal

Process involves changing amplitude/frequency/phase of carrier

This process is called modulation

Calculate B = μ₀nI where n = 4000 turns/m

Energy density = B²/2μ₀

Substitute and solve: (μ₀²n²I²)/2μ₀ = 3.2π J/m³

Use relation: B = BH/cosθ where BH = 3.0G and θ = 30°

Substitute values: B = 3.0/cos30°

Calculate: B = 3.0/0.866 = 3.5G

Apply Curie's law: x = C/T where C is Curie constant

Write equations: x₁ = C/T₁ and x₂ = C/T₂

Cross multiply to get x₁T₁ = x₂T₂

Use cyclotron principle: ω = qB/m (constant)

Linear velocity v = ωr

As r increases, v increases but ω stays constant

Consider magnetic field direction inside solenoid (along axis)

Analyze force on moving charge: F = qv×B

When v parallel to B, F = 0, no deflection

Consider signal flow

Starts with information source

Ends with user of information

Understand definition of magnetic susceptibility

For ferromagnetic materials, χ = M/H >> 1

Materials strongly attracted to magnetic field have χ >> 1

Use formula T = 2π√(I/MB) for time period

Calculate single oscillation period: T = 2π√[(12×10⁻⁶)/(6×10⁻²×2×10⁻²)] = 0.6 s

For 20 oscillations: Total time = 20×0.6 = 12 s

Analyze each statement against magnet properties

Statement D contradicts basic magnet property

Bar magnets always produce magnetic fields

Use magnetic field formulas: B_center = μ₀I/2R and B_axis = μ₀IR²/2(R²+x²)³/²

Set up ratio equation: B_center/B_axis = 5√5

Solve for x = 0.2m

Use mv² = qBr and KE = ½mv²

For same KE, v ∝ 1/√m

Radius ratio depends on √m/q

Use range formula: Range = √(2Rh)

Convert units to meters

Calculate: √(2×6.4×10⁶×500) = 80 km

Consider factors affecting Earth's field: core composition, latitude, altitude

Note that field lines are not uniform

Recognize variation due to geographical location

Understand magnetic properties: Ferromagnetic (strong attraction), Paramagnetic (weak attraction), Diamagnetic (weak repulsion)

Identify behavior of each needle based on material

Determine combined effect when magnet approaches

Use relation between V and velocity: ½mv² = qV

Use magnetic force equation: qvB = mv²/r

Combine equations and solve for new V

Use mv²/r = qvB for circular motion

Note that velocity remains constant when B changes

Calculate new radius using new B value

Recall vector form of Biot-Savart law

Note that r is unit vector in direction of position vector

Check dimensions of each option

B must be perpendicular to both E and direction of propagation

Magnitude B = E/c

Phase must be same as E field

Magnetic field needed = Coercivity = 3x10³ Am⁻¹

For solenoid, H = nI where n = 1000 turns/m

Solving 3x10³ = 1000I gives I = 3A

Wavelength λ = c/f where c = 3x10⁸ m/s

At 5MHz, λ = (3x10⁸)/(5x10⁶) = 60m

Antenna length = λ/4 = 15m

Magnetic domains contain groups of atoms with aligned magnetic moments

At room temperature, thermal energy prevents perfect alignment of all domains

Domains are partially aligned, giving net magnetization less than maximum possible

Work done = -M·B(cosθ₂ - cosθ₁) where θ₁ and θ₂ are initial and final angles

For rotation about axis perpendicular to loop plane, angle between M and B doesn't change

Therefore cosθ₂ = cosθ₁ and work done = 0

Current sensitivity = NAB/k where N is number of turns. When N is tripled, current sensitivity triples

Voltage sensitivity = NAB/Rk where R also increases by factor of 3 (more turns means more resistance). N/R ratio remains constant

Therefore only current sensitivity changes, not voltage sensitivity

Permanent magnets consist of magnetic domains where groups of atomic magnetic moments are aligned in the same direction. At room temperature, thermal energy causes some randomness in alignment.

Perfect alignment is impossible at room temperature because thermal energy causes some domains to be misaligned. This is why heating a magnet reduces its magnetization.

The domains maintain partial alignment, which gives the magnet its permanent magnetic properties while still being affected by thermal agitation.

Magnetic field at center B = μ₀NI/2r. Magnetic moment M = NIA where A = πr². Initial ratio x = B/M.

When current (I) doubles and radius (r) doubles: New B = μ₀N(2I)/2(2r) = B/2. New M = N(2I)π(2r)² = 4M.

New ratio = (B/2)/4M = x/8. Both changes affect ratio.

Use Curie law: M ∝ B/T where M is magnetization, B is magnetic field, T is temperature.

Set up proportion: M₁/M₂ = (B₁/T₁)/(B₂/T₂). Substitute M₁=8, B₁=0.6, T₁=4, B₂=0.2, T₂=16.

Calculate: M₂ = 8 × (0.2/16)/(0.6/4) = 2/3 Am⁻¹

In cyclotron, KE = q²B²r²/2m where q is charge, B is magnetic field, r is radius, m is mass.

Compare q/m ratios: Proton (1/1), Deuteron (1/2), α-particle (2/4=1/2). For same B and r, KE ∝ (q/m)².

Deuteron has smallest q/m ratio, hence gains minimum KE.

Biot-Savart law gives magnetic field due to current element. The field depends on current (i), length element (dl), and varies inversely as cube of distance (r³).

Vector nature is important: field is perpendicular to both current element and position vector (cross product). Direction follows right-hand rule.

Final expression is μ₀idl×r/4πr³, where × indicates cross product. Field decreases as cube of distance.

Earth's field has horizontal and vertical components

At poles, field lines are vertical

Therefore, horizontal component is zero only at magnetic poles

Particle needs centripetal force from magnetic force: qvB = mv²/r

On axis, B = μ₀nI. Maximum speed occurs when radius of curvature equals solenoid radius

Solve for v: v = μ₀nIqr/2m

Use Ampère's law for force between parallel currents: F/L = μ₀I₁I₂/2πr

Insert values: μ₀ = 4π×10⁻⁷, I₁=I₂=10A, r=0.1m

F/L = 2×10⁻⁴N/m, attractive for same direction currents

Magnetic force F = qv×B requires velocity

For stationary electron, v = 0

Therefore F = 0, electron remains stationary

Let's verify each value systematically. For B₀, we use E₀/B₀ = c. Given E₀ = 120 N/C, B₀ should be 120/3×10⁸ = 400nT. This is correct.

For angular frequency ω = 2πf = 2π(50×10⁶) = π×10⁸ rad/s. This matches given value. For wavelength λ = c/f = 3×10⁸/(50×10⁶) = 6m, also correct.

The propagation constant k = ω/c = (π×10⁸)/(3×10⁸) = π/3 rad/m, not 2.1 rad/m. This is the incorrect value.

Let's think about the electromagnetic spectrum order: gamma rays, X-rays, UV, visible, IR, microwaves, radio waves.

Wavelength increases as we go down this list. Energy decreases as wavelength increases (E=hc/λ).

Microwaves have longer wavelength than IR, UV, and X-rays. This makes them perfect for communication and heating!

Let's think about cosmic rays - they're charged particles. Charged particles are deflected by magnetic fields.

Earth's magnetic field is strongest at poles. If cosmic rays are deflected more at poles, this proves field exists!

Other options don't prove field: rotation doesn't create field, iron content alone doesn't prove active field, ionosphere is separate phenomenon.

The magnetic field along axis of circular coil follows special formula: B = μ₀nIr²/[2(x² + r²)³/²], where x is distance from center.

Here x = 3r. Substitute this: B = μ₀nIr²/[2(9r² + r²)³/²] = μ₀nIr²/[2(10r²)³/²]

Simplify carefully: r² terms cancel appropriately, giving B = μ₀nI/16r

For a finite solenoid, the field at the end is half of the field at center. Let's first find field at center: B = μ₀nI, where n is turns per unit length.

Calculate n = 100/0.5 = 200 turns/m. Then B_center = (4π×10⁻⁷)(200)(2.5)

At the end, B = B_center/2 = 3.14×10⁻⁴ T. Always check units match!

Use Curie law: χ ∝ 1/(T-Tc)

For Co: χ₁ ∝ 1/(1600-1400) = 1/200, For Fe: χ₂ ∝ 1/(1600-1000) = 1/600

Ratio χ₁/χ₂ = 600/200 = 3

For EM waves, E/B = c

c = 3×10⁸ ms⁻¹

Verify units and magnitude

For circular motion in B field: r = mv/qB

Compare mp/qp : mα/qα ratios

rp:rα = (mp/e):(4mp/2e) = 1:2

BH = Bcosθ where θ = dip angle

BH = 3×10⁻⁵ = Bcos45°

B = 3×10⁻⁵/cos45° = 3√2×10⁻⁵T

Torque τ = M×B = MBsinθ

When θ = 0°, sinθ = 0

Therefore τ = 0 at θ = 0°

For paramagnetic materials, susceptibility χ∝1/T (Curie's law)

Set up ratio χ₁/χ₂=T₂/T₁

Calculate χ₂=(300/200)×1.2×10⁻⁵=1.8×10⁻⁵

Use B=μ₀IR²/2(R²+x²)³/² for field at axial point

Compare with centre field B₀=μ₀I/2R and set ratio to 64

Solve (R²+x²)³/²=64R³ to get x=R√15

Consider electron as a moving charge

Moving charge produces electric field due to its charge

Also produces magnetic field due to its motion

Understand that hysteresis means lagging of magnetic induction B behind magnetizing force H

Only ferromagnetic materials show retention of magnetization after removing external field

This retention causes hysteresis loop formation, unique to ferromagnetic materials

Use Biot-Savart law: B∝NI/R where N is number of turns

For same current and radius, B∝N

B₂/B₁=3/9=1/3, so B₂=B₁/9

Calculate time period T=2πm/qB=2π/2=π s

Find angle of deviation θ=ωt=2πt/T=π/6=30°

Resolve velocity components: v=10(cos30°i-sin30°j)=5(-√3i+j)