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Correct choice is (c) 3

To explain: At the midpoint ➔ CASE 1:

B1=\(\frac {\mu_o I_1}{2\pi d} – \frac {\mu_o I_2}{2\pi d}\)

B1=\(\frac {\mu_o (I_1 – I_2)}{2\pi d}\)=50 μT

At the midpoint ➔ CASE 2:

B2=\(\frac {\mu_o I_1}{2\pi d} + \frac {\mu_o I_2}{2\pi d}\)

B1=\(\frac {\mu_o (I_1 + I_2)}{2\pi d}\)=100 μT

\(\frac {B1}{B2}=\frac {I_1-I_2}{I_1+I_2}=\frac {50 \mu T}{100 \mu T}\)

\(\frac {I_1-I_2}{I_1+I_2}=\frac {1}{2}\)

Using componendo and Dividendo RULE:

\(\frac {(I_1+I_2) \, + \, (I_1-I_2)}{(I_1-I_2) \, – \, (I_1+I_2)}=\frac {1+2}{1-2}\)

\(\frac {I_1}{-I_2}=\frac {3}{-1}\)

Therefore, \(\frac {I_1}{I_2}\)=3

The correct option is (d) 3B

For explanation: Firstly, the magnetic FIELD at the MIDPOINT due to current in AB:

BAB=μo×\(\frac {2 \TIMES 4}{4\pi d}=\frac {8\mu_o}{4\pi d}\) ………………..1

Magnetic field at midpoint due to current in CD:

BCD=μo×\(\frac {2 \times 3}{4\pi d}=\frac {6\mu_o}{4\pi d}\) …………………..2

The NET magnetic field = BAB – BCD ➔ 1 – 2

B = \(\frac {2 \mu_o}{4\pi d}\) ……………………….3

Now, when 4A is switched off, then the magnetic field at the midpoint will be due to the current in CD, i.e. due to 3A current

B^‘ = μo×\(\frac {2 \times 3}{4\pi d}=\frac {6\mu_o}{4\pi d}\) ……………..4

Comparing 3 and 4:

Therefore, B^‘ = 3B

The correct choice is (a) F=\(\frac {\mu_o}{4\pi } \times \frac {2I_1 I_2}{r}\) × l

The best I can EXPLAIN: If TWO LINEAR current carrying conductors of unequal length are held parallel to each other, then the force on a LONG conductor is due to MAGNETIC field interaction due to currents of short conductor and long conductor. The force on long conductor is force to that on the short conductor given by:

F=\(\frac {\mu_o}{4\pi } \times \frac {2I_1 I_2}{r}\) × l

Right choice is (a) Neutron

Easy explanation: Neutrons, being electrically NEUTRAL cannot be accelerated in a cyclotron. The cyclotron is a heavy PARTICLE accelerator USED to accelerate charged particles LIKE protons, deuterons and α-particles to high VELOCITIES.

Right option is (a) θ = 90^o

For explanation: When θ = 90^o ➔ τ = NIBA.

The torque is maximum when the PLANE of the LOOP is parallel to the MAGNETIC field. So the ANGLE θ has to be 90^o.

The correct choice is (b) By connecting a low RESISTANCE shunt in PARALLEL to the GALVANOMETER

Easiest explanation: A galvanometer can be converted into an ammeter of given range by connecting SUITABLE low resistance S CALLED shunt in parallel to the given galvanometer, whose value is given by:

S = \( [ \frac {I_g}{(I – I_g)} ] \)G

Where Ig is the current for full scale deflection of galvanometer, I is the current to be measured by the galvanometer and G is the resistance of galvanometer.

The correct OPTION is (a) True

The best I can explain: Yes, voltage and current sensitivity are RELATED to each other. Current sensitivity

IS = \(\frac {NAB}{k}\); Voltage sensitivity = \(\frac {\theta }{V} = \frac {\theta }{IR} = \frac {NAB}{Kr}\)

THEREFORE, voltage sensitivity ➔ VS = \((\frac {1}{R})\) x IS.

The CORRECT answer is (b) \(\frac {-3F}{5}\)

The EXPLANATION is: First case:

F=\(\frac {\mu_o}{2\pi } \big [ \frac {I_1 I_2}{r} \big ] \)L

Second case:

F^‘=\(\frac {\mu_o}{2\pi } \big [ \frac {(-3I_1)(I_2)}{5R} \big ] \)l

\(\frac {F^{‘}}{F} = \frac {\frac {\mu_o}{2\pi } \big [ \frac {(-3I_1)(I_2)}{5r} \big ] l}{\frac {\mu_o}{2\pi } \big [ \frac {I_1 I_2}{r} \big ] l}\)

\(\frac {F^{‘}}{F}= -\frac {3}{5}\)

Therefore, F^‘=\((-\frac {3}{5})\)F=\(\underline{\frac {-3F}{5}}\)

The CORRECT answer is (d) The shape of the loop

For explanation I would say: Torque on a planar current loop DEPENDS upon current, the strength of the MAGNETIC FIELD, area of the loop and the orientation of the loop in the magnetic field. It is independent of the shape of the loop.

Right choice is (a) True

Easy explanation: Yes, in the magnetic field at the center in toroid is stronger than that in a solenoid. This is due to its ring STRUCTURE. The magnetic field INSIDE and OUTSIDE the toroid is ZERO. Toroid does not have a uniform magnetic field inside it unlike a solenoid.

The correct option is (b) 0.125 Ω

Best explanation: GIVEN: Ig = 0.05 A; G = 50 Ω; I = 20 A

Required EQUATION ➔ S = Ig × \(\frac {G}{(I – I_g)}\)

S = \(\frac {0.05 \times 50}{(20 – 0.05)}\)

S = \(\frac {2.5}{19.95}\)

S = 0.125 Ω

Therefore, the required SHUNT resistance is 0.125 Ω.

The correct OPTION is (C) 75 Nm

Explanation: Torque = NIBA cosθ [Angle between field and plane of coil].

Torque = 500 × 30 × 0.5 × 200 × 10^-4 × cos (60)

Torque =75 Nm.

Therefore, the torque is 75 Nm.

Correct answer is (c) 1 x 10^-5 T

For explanation I would say: Given: I = 100A; r = 2 m; \(\frac {\mu_o}{4 \pi }\) = 10^-7

The required equation → B = \((\frac {\mu_o}{4 \pi }) \, \times \, \frac {2I}{r}\)

B = 10^-7 × 2 × \(\frac {100}{2}\)

B = 10^-7 × 100

B = 1 × 10^-5 T

Therefore, the magnitude of the MAGNETIC FIELD is 1 × 10^-5 T and the direction of the magnetic field is south.

Correct CHOICE is (c) ZERO

To EXPLAIN I would say: Work done is zero. Since a magnetic field exerts a force perpendicular to the direction of motion of the charged particle, no work is done by it on the charged particle. THEREFORE, the total work done by the magnetic field during the time the particle is trapped is zero.

Correct choice is (b) Straight PATH

To elaborate: The electron will continue to follow its straight path because a parallel magnetic FIELD does not exert any force on the electron. So, there won’t be any change in its TRAJECTORY when the electron enters a uniform magnetic field.

Correct ANSWER is (d) 30 Ω

Easy explanation: GIVEN: G = 10 Ω; V = 2 V; Ig = 0.05 A

The REQUIRED equation ➔ Ig = \(\frac {V}{(R + G)}\)

0.05 = \(\frac {2}{(R + 10)}\)

(R + 10) = \(\frac {2}{0.05} = \frac {200}{5}\) = 40

R = 40 – 10

R = 30 Ω

Right choice is (b) \(\frac {5\mu_0}{6}\)

Easiest explanation: Since the TWO coils are concentric and in the same plane, carrying currents in opposite directions, the total magnetic field at the center of the concentric coils is given by:

B = B1 – B2 = \(\frac {\mu_0}{4\PI } [ \frac {2\pi N_1 I_1}{r_1} ] \, – \, \frac {\mu_0}{4 \pi } [ \frac {2 \pi N_2 I_2}{r_2} ]\)

B = \(\frac {\mu_0}{2} [ \frac {N_1 I_1}{r_1} \, – \, \frac {N_2 I_2}{r_2} ]\)

B = \(\frac {\mu_0}{2} \big [ \)20 × \( \frac {0.4}{0.6} \) – 20 × \( \frac {0.5}{0.8} \big ] \)

B = \(\frac {\mu_0}{2} [ \frac {40}{3} \, – \, \frac {25}{2} ]\)

B = \(\frac {\mu_0}{2} [ \frac {80-75}{6} ] \)

B = \(\frac {5\mu_0}{6}\)

The correct answer is (c) 8 × 10^4 A/m

The explanation: Given: number of TURNS (n) = 200 turns per meter; CURRENT (I) = 2 A; RELATIVE permeability (µr) = 200

Magnetic INTENSITY is given by: H = nI = 200 × 2 = 400 A/m

μr = 1 + γ → γ is the magnetic susceptibility of the material

γ = μr – 1

Magnetization → M = γ x H

M = (μr – 1) × H

M = (200 – 1) × 200

M = 199 × 200

M = 79600

M = 7.96 × 10^4A/m ≈ 8 × 10^4 A/m

The correct OPTION is (d) 1.4 × 10^-9 T

The best I can explain: dB = \(\frac {(\mu_0 I dl sin (\THETA))}{(4\pi r^2)}\).

dB = \(\frac {(4 \pi \, \TIMES \,10^{-7} \, \times \,8 \, \times \,1 \, \times \,10^{-2} \, \times \,sin 45^o}{(4 \pi \, \times \,22)}\)

dB = 1.4 × 10^-9 T.

The correct answer is (a) 5.25 × 10^-8 s

The EXPLANATION is: Time PERIOD, t = \(\frac {\pi m}{qB}\)

t = \(\frac {(3.14 \, \times \, 6.68 \, \times \, 10^{-27})}{(3.2 \, \times \, 10^{-19} \, \times 1.25)}\)

t = 5.25 × 10^-8 s.

Correct choice is (d) 1.875 × 10^11 C/Kg

The best explanation: Radius, R=\(\FRAC {MV}{eB}\).

\(\frac {e}{m} = \frac {v}{rB}\)

\(\frac {e}{m} = \frac {3 \, \TIMES \, 10^7}{(8 \, \times 10^{-3} \, \times \, 2 \, \times \, 10^{-2})}\)

\(\frac {e}{m}\) = 1.875 × 10^11 C/Kg.

Right choice is (B) False

Explanation: If θ = 90^o → sin θ = 1

→ Then DB is maximum.

The magnetic FIELD due to a current ELEMENT is maximum in a PLANE passing through the element and perpendicular to its axis.

Correct choice is (c) 1.56 T

To explain: Cyclotron frequency = \(\frac {qB}{2 \pi m}\).

B = \(\frac {(2 \pi mf_c)}{(q)}\)

B = \(\frac {(2 \, \TIMES 3.142 \, \times 3.3 \, \times \, 10^{-27} \, \times \, 12 \, \times \, 10^6)}{(1.6 \, \times \, 10^{-19})}\)

B = 1.56 T

Right option is (B) False

Explanation: When a CHARGE moves parallel or ANTIPARALLEL to the DIRECTION of the magnetic field, it EXPERIENCES a minimum (zero) force.

When θ = 0^o or 180^o,

Fm = qvBsinθ = qvB (0) = 0.