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The correct option is (B) \(\frac {\mu_o 2I^3}{4πb}\)

Explanation: The expression for force between 2 parallel CURRENT carrying conductors is given by:

f=\(\frac {\mu_o}{4\pi } \big [ \frac {I_1 I_2}{b} \big ] \)

In this CASE, the current passing through both of them are the same, i.e. I1=I2

Therefore, the force per unit LENGTH is given as:

f=\(\frac {\mu_o}{4\pi } \big [ \frac {I^2}{b} \big ] \)

The correct answer is (b) The electron beams will ATTRACT each other and the conductors also attract each other

The BEST I can explain: Since, both the CURRENT carrying conductors are moving in the same direction, they will attract each other. Moving electron beams is equivalent to an electric current in opposite direction. Therefore, there will ATTRACTION between the conductors X and Y as well as between the electron beams P and q.

The correct choice is (c) 1.893 × 10^-5 N m

To explain: Given: μ0 = 4 × 3.14 × 10^-7; number of turns of SOLENOID = 100; Current PASSING through the solenoid = 5 A; Number of turns of coil = 20; Current passing through the coil = 0.6A; Length of the solenoid = 0.5 m; Radius of coil = 0.02 m

Magnetic field of solenoid (B) = μ0nI = μ0 × \(\FRAC {100}{0.5}\) × 5 = 4 × 3.14 × 10^-7 × \(\frac {100}{0.5}\) × 5

Magnetic moment of the coil (M) = I × A × N = 0.6 × (0.02)^2 × 20

Torque (τ) = MBsin (90^o)

τ = 0.6 × (0.02)^2 × 20 × 4 × 3.14 × 10^-7 × \(\frac {100}{0.5}\) × 5[sin (90^o) = 1]

τ = 1.893 × 10^-5 N m

Right answer is (b) False

Explanation: Force on a CHARGE ‘q’ MOVING with a velocity ‘V’ in a MAGNETIC field is given by:

 F = qvB sinθ.

The direction of the force is given by FLEMING’s left-hand rule.

The CORRECT choice is (c) The principle of superposition does not apply to both

Easy explanation: The principle of superposition applies to both fields. This is because the magnetic field is LINEARLY related to its source, NAMELY, the current ELEMENT and the ELECTROSTATIC field is related linearly to its source, the electric charge.

Correct option is (a) F = 0

For explanation I would say: Magnetic FORCE on a proton is Fm= qvBsin(θ).

A proton moving parallel or ANTIPARALLEL to a magnetic field does not EXPERIENCE any force. Hence, the proton does not get deflected while moving through a magnetic field REGION.

Correct ANSWER is (a) The force between TWO parallel CURRENT carrying wires is independent of the radii of the wires

For explanation I would say: The force between two parallel current carrying wires is independent of the radii of the wires.

From Ampere’s circuital law, the MAGNITUDE of the field due to the first conductor can be given by:

B = \(\frac {\mu_o I_1}{2\pi d}\)

The force between the parallel plates is given by:

B= \([\frac {\mu_o I_1 I_2}{2\pi d}] \) L

From this equation, we understand that the force between the parallel current carrying wires does not depend on the ‘radii’.

The correct option is (a) True

For explanation I would say: Yes, when infinitely LONG PARALLEL wires CARRY equal currents in the same direction, the magnetic field at the midpoint in between the two wires is zero. This is because, when the wires carry equal currents in the same direction, the magnetic field created at the midpoint will CANCEL out each other and THUS, the field is zero.

Right answer is (d) Zero

The explanation is: As the magnetic FORCE acts in a direction perpendicular to the direction of the velocity or the direction of MOTION of the charged particle, so the work DONE is zero.

W = F × dl × cos 90^°

W = 0.

Correct choice is (b) False

The explanation: An ammeter is a low resistance instrument and it is ALWAYS connected in series to the circuit. An ideal ammeter has zero resistance. Voltmeter is a high resistance instrument and it is always connected in parallel with the circuit ELEMENT ACROSS which potential DIFFERENCE is to be measured. An ideal voltmeter has INFINITE resistance.

The correct option is (a) True

For explanation: Magnetic field at the CENTER of a circular arc of radius R, carries a current I and MAKES an angle θ at the center is GIVEN by

B = \(\FRAC {\mu_o I \theta}{4 \pi R}\)

In this case θ = \(\frac {\pi }{4}\)

B = \(\frac {\mu_o I (\frac {\pi }{4})}{4 \pi R}\)

B = \(\frac {\mu_o I}{16 R}\)

Correct answer is (d) 1.50 × 10^-2 Wb/m^2

To explain I would say: The magnetic field induction at the center of a long solenoid is B = μonI→B ∝ NIN = number of turns per unit length of solenoid. I = current PASSING through the solenoid.

\(\frac {B1}{B2} = \frac {n1}{n2} [ \frac {I1}{I2} ] \)

\(\frac {B1}{B2} = \frac {500}{300} \, \TIMES \, \frac {I}{(\frac {I}{3})}\)

\(\frac {B1}{B2}\) = 5

B2 = \(\frac {B1}{5}\)

B2 = \(\frac {7.54 \, \times \, 10^{-2}}{5}\)

B2 = 0.01508 = 1.50 × 10^-2Wb/m^2

The correct answer is (C) 5 div/mA, 0.167 div/mV

To explain: N = 50; A = 70 mm^2; B = 0.1 T; k = 7 × 10^-8 N m/rad; R = 30 Ω

Current sensitivity (IS) = \(\frac {NAB}{k}\) = 50 × 70 × 10^-6 × \(\frac {0.1}{7}\) × 10^-8

= 350 × \(\frac {100}{7}\) = 5 × 10^3 div/amp

= 5 div/mA

Voltage sensitivity (VS) = \(\frac {NAB}{KR} = \frac {I_S}{R}\)

= \(\frac {5}{30} = \frac {1}{6}\)

= 0.167 div/mV

The correct option is (c) Zero

The best explanation: The north POLE of a magnet will not experience any force. This is because a STATIONARY charge does not PRODUCE any magnetic field. Therefore, the force EXPERIENCED by the magnet at the POLES is zero.

The correct option is (a) B = \( [ \frac {\mu_o I}{4 \pi r} ] \int_0^{2\pi r} \)dlsin⁡90

Best explanation: The magnetic field strength at any point at center of CIRCULAR LOOP carrying current I and radius r is given as:

B = \( [ \frac {\mu_o I}{4 \pi r} ] \int_0^{2\pi r} \)dlsin⁡90

It can also be expressed as → B = \( [ \frac {\mu_o I}{4 \pi r} ] \) × 2πror B = \( [ \frac {\mu_o I}{2 r} ] \) DIRECTION.

It is inwards if the current is FLOWING in the clockwise direction and it is onwards if the current is flowing in the anticlockwise direction.

Correct option is (B) 5 × 10^-2 N m

The best explanation: The magnitude of torque experienced by a current carrying coil kept in a UNIFORM magnetic field is given by:

τ = MB sin (θ) = (NIA) × B sin (θ)

M → Magnetic moment of the coil

N → NUMBER of turns

I → Current in coil

A → Area of the coil

θ → Angle between NORMAL to the plane of the coil and the direction of the magnetic field

Given: N = 50; A = 4 × 5 = 20 cm^2 = 20 × 10^-4 m^2; B = 0.5 T; I = 2 A

θ = 90^o – 60^o = 30^o

τ = (NIA) × B sin (θ)

τ = 50 × 2 × 0.5 × 20 × 10^-4 × sin (30^o)

τ = 0.05

τ = 5 × 10^-2 N m

Correct CHOICE is (b) Tesla

For EXPLANATION: The SI UNIT of the magnetic field is the tesla (T). One tesla is 10^7 times the magnetic field produced by a conducting wire of length one METER and carrying a current of one AMPERE at a distance of one meter from it and perpendicular to it.

Right ANSWER is (d) α-particle

The best explanation: F = qvB.

For α-particle, Q = 2e, Fα = 2evB

For β-particle, q = e, Fβ = evB.

Thus, the α-particle will EXPERIENCE maximum FORCE.

The correct choice is (c) S = \(\frac {G}{(n – 1)}\)

For EXPLANATION I WOULD say: In order to increase the range of an AMMETER n times, the value of shunt resistance to be connected in parallel is given by:

S = \(\frac {G}{(n – 1)}\)

In order to increase the range of voltmeter n times, the value of resistance to be connected in series with GALVANOMETER is given by:

R = (n – 1) G

Correct choice is (C) Am

Easy explanation: The SI unit of magnetic pole strength is ampere meter (Am).

The strength of a magnetic pole is said to be ONE-ampere meter if it repels an EQUAL and similar pole with a force of 10^-7 N when PLACED in vacuum at a distance of one meter from it.

Correct answer is (d) Current sensitivity is expressed as the EXACT reverse of the galvanometer CONSTANT

Easiest explanation: Current sensitivity is defined as the deflection PRODUCED in the galvanometer, when unit current flows through it.

IS = \(\frac {NAB}{k}\)………………1

The unit of current sensitivity is rad A^-1 or DIV A^-1.

From 1, we can UNDERSTAND that current sensitivity is expressed as the exact reverse of galvanometer constant (G = \(\frac {k}{NAB}\)).

Right answer is (a) B = μ0 NI

For explanation: AMPERE’s circuital law:

∮ B.dl = ∮ B.dlcos⁡0 = B ∮ dl = B.2πr…………………..1

But, for a toroid:

∮ B.dl = μ0 × total CURRENT THREADING the toroid

∮ B.dl = μ0 × total number of turns in the toroid × I

∮ B.dl = μ0 N2πrI ……………………………2

From 1 and 2

B.2πr = μ0 N2πrI

Therefore, B = μ0NI

Correct answer is (b) 393 μT

For explanation: Field ALONG axis of coil → B =\( \frac {\mu_o IR^2}{2 \BIG ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] } \)

At the coil of the coil → B1 = \( \frac {\mu_o I}{2R}\)

\(\frac {B_1}{B} = \frac {\mu_o I}{2R} \, \times \, \frac {2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] }{\mu_o IR^2} = \frac {2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] }{R^3}\)

B1 = \(\frac {48 [4^2 + 7^2]^{\frac {3}{2}}}{4^3}\)

B1 = \(\frac {48 \times (65)^{\frac {3}{2}}}{4^3}\)

B1 = 131.01 × 3

B1 = 393.03 = 393 μT

Right answer is (a) [M L T^-2 A^-2]

The explanation: MAGNETIC permeability = Magnetic flux density × [Magnetic FIELD STRENGTH]^-1.

μ = [M L^0 T^-2 A^-1] × [M^0 L^-1 T^0 A^1]^-1

μ = [M L T^-2 A^-2].

The correct answer is (c) 2

Easiest EXPLANATION: The magnetic field due to a long straight wire of radius x carrying a current I at a point distant r from the AXIS of the wire is GIVEN by:

Bin = \(\frac {\mu_o IR}{2\pi x^2}\)(r < x)

Bin = \(\frac {\mu_o I}{2\pi r}\)(r > x)

The magnetic field at a distance of r = \(\frac {x}{4}\):

B1 = \(\frac {\mu_o I (\frac {x}{4})}{2\pi x^2} = \frac {\mu_o I}{8\pi x}\)

The magnetic field at a distance of r = 8x:

B2 = \(\frac {\mu_o I}{2\pi (8x)} = \frac {\mu_o I}{16\pi x}\)

Therefore, \(\frac {B1}{B2} = \frac {\frac {\mu_o I}{8\pi x}}{\frac {\mu_o I}{16\pi x}}\)

\(\frac {B1}{B2} = \frac {16}{8}\) = 2

Correct answer is (d) Biot-Savart LAW

The explanation: In classical electrodynamics, the magnetic FIELD GIVEN by a CURRENT loop and the ELECTRIC field caused by the corresponding dipoles in sheets are very similar, as far as we are far away from the loop, which enables us to deduce Ampere’s magnetic circuital law from the Biot-Savart law easily.

Right answer is (C) T A^-1 m

To elaborate: Magnetic PERMEABILITY of free space is a MEASURE of the amount of resistance ENCOUNTERED when forming a magnetic field in a classical vacuum. The SI unit of permeability is weber ampere^-1 metre^-1 (Wb A^-1 m^-1) or tesla ampere^-1meter (T A^-1 m).

The correct option is (b) Speed of the particle

Easiest explanation: The radius of the circular path of the CHARGED particle increases in direct PROPORTION to its speed. CONSEQUENTLY, both its time-period and frequency of revolution are INDEPENDENT of its speed. So, the ‘X’ is the speed of the particle.

The correct answer is (a) θ = 90^o

The explanation is: When θ = 90^o→ Fmis MAXIMUM.

Thus a charge EXPERIENCES a maximum force when it moves perpendicular to the direction of the magnetic FIELD. So, this is the condition when the force EXPERIENCED is maximum.

The CORRECT choice is (b) 50 Ω

Easy explanation: GIVEN: V = 10 V; Ig = 0.010A; R = 950 Ω

Required equation ➔ R = \((\frac {V}{I_g})\) – G ➔ G = \((\frac {V}{I_g})\) – R

G = \((\frac {10}{0.010})\) – 950

G = 1000 – 950

G = 50 Ω

Therefore, the resistance of galvanometer is 50 Ω.

Correct option is (b) False

Explanation: The FACTOR \((\frac {EH}{4\pi m_e})\) is the least value of the magnetic MOMENT. This natural unit of magnetic moment is CALLED Bohr magneton. The value of Bohr magneton is 9.27 × 10^-24 Am^2. So, that MAKES the above statement false.

Right choice is (B) θ = 180^o

The EXPLANATION: When θ = 0^o or 180^o,

Fm= qvBsinθ = QVB (0) = 0.

So when a charge MOVES parallel or antiparallel to the DIRECTION of the magnetic field, it experiences a minimum force.

The CORRECT answer is (c) 1.63 × 10^-8 T

For explanation I WOULD SAY: B = \(\frac {MV}{qr}\).

B = \(\frac {(1.67 \, \times \, 10^{-27} \, \times \, 10^7)}{(1.6 \, \times \, 10^{-19} \, \times \, 6.4 \, \times \, 10^6)}\)

B = 1.63 × 10^-8 T

Therefore, the minimum magnitude of magnetic field should be 1.63 × 10^-8 T.

Correct answer is (C) B=\( \frac {\mu_o IR}{4 \pi \big ( (r^2 \, + \, x^2)^{\frac {3}{2}} \big ] } \)× 2πr

Best explanation: The EXPRESSION for MAGNETIC field on the axis of circular loop is given as:

B=\( \frac {\mu_o Ir}{4 \pi \big ( (r^2 \, + \, x^2)^{\frac {3}{2}} \big ] } \)× 2πr

It is towards the loop if current in it is in clockwise direction and it is away from the loop if current in it is in anticlockwise direction.

The correct answer is (b) 2.34 × 10^-8 s

The best I can EXPLAIN: Time period, t = \(\FRAC {\pi m}{qB}\)

t = \(\frac {(3.14 \, \TIMES 1.67 \, \times \, 10^{-27})}{(1.6 \, \times \, 10^{-19} \, \times \, 1.4)}\)

t = 2.34 × 10^-8 s

The CORRECT option is (c) Measurement of small currents

The explanation is: Moving Coil GALVANOMETER is an INSTRUMENT used for the detection and measurement of current. It is sensitive instrument and can measure current even if it is only a few microamperes. It was INVENTED by Johann Schweigger in the 1800s.

Correct ANSWER is (b) 2 × 10^-11 N

For explanation I WOULD say: F = qvB sinθ.

F = 1.6 × 10^-19 × 5 × 10^7 × 5 × sin30^o

F = 2 × 10^-11 N.

Correct option is (a) Zero

The best EXPLANATION: A stationary charge does not PRODUCE any magnetic field and it does not suffer any interaction against the external magnetic field. HENCE the FORCE EXERTED is zero.

Correct ANSWER is (b) TESLA

For explanation I would say: The SI unit of the magnetic field is tesla, named after the great SCIENTIST Nikola tesla. 1 tesla is 10^7 TIMES the magnetic field produced by a conducting wire of length one metre and carrying a current of one ampere at a distance of one metre from it and perpendicular to it.

The correct answer is (c) Left

Explanation: According to FLEMING’s left-hand rule, the magnetic FORCE will act TOWARDS left. When the POSITIVE charge enters a region of magnetic field directed towards north, the magnetic force will act towards left.

The correct choice is (d) By connecting a high resistance multiplier in series with the galvanometer

For explanation I WOULD say: A galvanometer can be CONVERTED into voltmeter of given range by connecting a high resistance called multiplier in series with the galvanometer, whose value is given as:

R = \((\frac {V}{I_g})\) – G

Where V is the voltage to be measured, IG is the CURRENT for full scale deflection of galvanometer and G is the resistance of galvanometer.

Correct option is (d) \(\FRAC {3G}{17}\)

Explanation: Ig = \((\frac {15}{100})\) × I

S = Ig × \(\frac {G}{I – I_g}\)

S = \( ( \frac {15}{100})\) × I × \(\frac {G}{(I – 15 \times \frac {I}{100})}\)

S = [0.15 × \( \frac {I}{(I – 0.15 \times I)} \) ] × G

S = [0.15 × \( \frac {I}{0.85 \times I} \) ] × G

S = \( \frac {15}{85} \) × G

S = \( \frac {3G}{17} \)

THEREFORE, the shunt resistance required is \( \frac {3G}{17} \).

The CORRECT answer is (b) 5.7 cm

The EXPLANATION is: Radius → r = \(\frac {mv}{eB}\)

r = \(\frac {(9.1 \, \times \, 10^{-31} \, \times \, 5 \, \times \, 10^7)}{(1.6 \, \times \, 10^{-19} \, \times \, 5 \, \times \, 10^{-3})}\)

r = 5.7 cm

Therefore, the radius of the path is 5.7 cm.

The correct OPTION is (b) False

The best I can explain: ELECTRONS cannot be accelerated in a cyclotron. A large increase in their energy increases their velocity to a very large extent. This throws the electrons out of step with the oscillating field. A cyclotron can only accelerate particles such as PROTONS, deuterons, and alpha – particles.

Right option is (a) θ = 0^o

For explanation I WOULD say: When θ = 0^o ➔ τ = 0.

The torque is minimum when the PLANE of the loop is perpendicular to the MAGNETIC FIELD, so the angle θ should be 0^o.