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NEET NEET Questions — Free MCQ Practice

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A 50 turns circular coil has a radius of 3 cms, it is kept in a magnetic field acting normal to the area of the coil. The magnetic field B increased from 0.10 tesla to 0.35 tesla in 2 milliseconds. The average induced emf in the coil is-

A.

1.77 volts

B.

17.7 volts

C.

177 volts

D.

0.177 volts

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Explanation

$2 . ϕ = NBA ϕ_{1} = 50 \times π \times {(3 \times 10^{- 2})}^{2} \times 0.1 = 141.3 \times 10^{- 4} Wb ϕ_{2} = 50 \times π \times {(3 \times 10^{- 2})}^{2} \times 0.35 = 494.5 \times 10^{- 4} Wb ∴ e = \frac{dϕ}{dt} = 17.7 volt$

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The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through the coil is approximately

A.

$0.1 μ_{0} Wb$

B.

$0.2 μ_{0} Wb$

C.

$1.0 μ_{0} Wb$

D.

$2.0 μ_{0} Wb$

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Explanation

$1 . L = \frac{Nϕ}{i} 8 \times 10^{- 3} = \frac{400 \times ϕ}{5 \times 10^{- 3}} \Rightarrow ϕ = \frac{40 \times 10^{- 6}}{400} Wb = 10^{- 7} Wb \Rightarrow ϕ = \frac{4 π \times 10^{- 7}}{4 π} Wb \Rightarrow ϕ = \frac{μ_{0}}{4 π} Wb \Rightarrow ϕ \approx 0.1 μ_{0} Wb$

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The current in an L – R circuit builds up to ${(3 / 4)}^{th}$ of its steady state value in 4 seconds. The time constant of this circuit is

A.

$\frac{1}{In 2} \sec$

B.

$\frac{2}{In 2} \sec$

C.

$\frac{3}{In 2} \sec$

D.

$\frac{4}{In 2} \sec$

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Explanation

$2 . I = I_{0} (l - e^{- t / τ}) where τ \to time constant ∴ \frac{3}{4} I_{0} = I_{0} (l - e^{- t / τ}) \Rightarrow \frac{3}{4} = l - e^{- t / τ} \Rightarrow e^{- t / τ} = \frac{1}{4} \Rightarrow \frac{- t}{τ} In e = In \frac{1}{4} \Rightarrow \frac{- 4}{τ} = - 2 In 2 \Rightarrow τ = \frac{2}{In 2}$

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The magnetic flux through each turn of a 100 turn coil is $(t^{3} - 2 t) \times 10^{- 3} Wb$ , where t is in second. The induced emf at t = 2 s is

A.

$- 4 V$

B.

$- 1 V$

C.

$+ 1 V$

D.

$+ 4 V$

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Explanation

$2 . ϕ = (t^{3} - 2 t) \times 10^{- 3} \frac{dϕ}{dt} = (3 t^{2} - 2) \times 10^{- 3} {\frac{dϕ}{dt}}_{t = 2} = (3 \times 4 - 2) \times 10^{- 3} Wb / s = 10^{- 2} Wb / s e = - N \frac{dϕ}{dt} = - 100 \times 10^{- 2} V = - 1 V$

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An emf of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = $\infty$ and at t = 1 second is -

A.

$\frac{e^{1 / 2}}{e^{1 / 2} - 1}$

B.

$\frac{e^{2}}{e^{2} - 1}$

C.

$1 - e^{- 1}$

D.

$e^{- 1}$

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Explanation

$2 . I = I_{0} (1 - e^{- Rt / L}) I_{0} = \frac{E}{R} (Steady current) When t = \infty I_{\infty} = \frac{E}{R} (1 - e^{- \infty}) = \frac{15}{10} = 1.5 I_{1} = 1.5 (1 - e^{- R / L}) = 1.5 (1 - e^{- 2}) ∴ \frac{I_{\infty}}{I_{1}} = \frac{1}{1 - e^{- 2}} = \frac{e^{2}}{e^{2} - 1}$

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A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self induction of the coil will be -

A.

$25 mH$

B.

$25 \times 10^{- 3} mH$

C.

$50 \times 10^{- 3} mH$

D.

$50 \times 10^{- 3} H$

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Explanation

$1 . ϕ = Li \Rightarrow NBA = Li Since magnetic field at the centre of circular coil carrying current is given by B = \frac{μ_{0}}{4 π} \times \frac{2 πNi}{r} ∴ N . \frac{μ_{0}}{4 π} . \frac{2 πNi}{r} . {πr}^{2} = Li \Rightarrow L = \frac{μ_{0} N^{2} πr}{2} Hence self inductance of a coil \frac{4 π \times 10^{- 7} \times 500 \times 500 \times π \times 0.05}{2} = 25 mH$

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A coil of radius 1 cm and turns 100 is placed in the middle of a long solenoid of radius 5 cm and having 8 turns/cm. The mutual induction in millihenry will be-

A.

0.0316

B.

0.063

C.

0.105

D.

Zero

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Explanation

1. Magnetic induction in the solenoid

$B = µ_{0} ni$

Magnetic flux linked with the coil

$ϕ = NAB = NA µ_{0} ni$

$M = \frac{ϕ}{i} = \frac{N A μ_{0} n i}{i} = {NAμ}_{0} n M = 100 \times π (1 \times 10^{- 2})^{2} \times 4 π \times 10^{- 7} \times 800 = 316 \times 10^{- 7} H = 0.0316 mH .$

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A copper rod of length 0.19 m is moving parallel to a long wire with a uniform velocity of 10 m/s. The long wire carries 5 ampere current and is perpendicular to the rod. The ends of the rod are at distances 0.01 m and 0.2 m from the wire. The emf induced in the rod will be-

A.

$10 μV$

B.

$20 μV$

C.

$30 μV$

D.

$40 μV$

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Explanation

$3 . The magnetic field at a distance x from the wire B (x) = \frac{μ_{0} i}{2 πx} EMF induced in an element of length dx at a distance x from wire = Bvdx ∴ Total EMF induced in the rod E = \int_{0.01}^{0.2} Bv dx = \int_{0.01}^{0.2} \frac{μ_{0} iv}{2 πx} dx = \frac{μ_{0} iv}{2 π} \int_{0.01}^{0.2} \frac{1}{x} dx E = \frac{μ_{0} iv}{2 π} {[\log_{e} x]}_{0.01}^{0.2} = \frac{μ_{0} iv}{2 π} [\log_{10} (0.2) - \log_{10} (0.01)] \times 2.303 E = \frac{4 π \times 10^{- 7} \times 5 \times 10}{2 π} [1.301] \times 2.303 = 2.99 \times 10^{- 5} V \approx 30 μV$

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A long solenoid having 1000 turns per cm is carrying alternating current of one ampere peak value. A search coil of area of cross-section $1 \times 10^{- 4} m^{2}$ and of 20 turns is placed in the middle of the solenoid so that its plane is perpendicular to the axis of the solenoid. The search coil registers a peak voltage $2.5 \times 10^{- 2} V$ . The frequency of the current in the solenoid is -

A.

1.6 per second

B.

0.16 per second

C.

15.9 per second

D.

15.85 per second

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Explanation

$4 . Flux linked with the search coil ϕ = {BAN}_{s} = μ_{0} {niAN}_{s} ∴ \frac{dϕ}{dt} = μ_{0} {nAN}_{s} \frac{di}{dt} i = i_{0} \sin ωt ∴ \frac{dϕ}{dt} = μ_{0} {nAN}_{s} i_{0} ω \cos ωt E_{\max} = {(\frac{dϕ}{dt})}_{\max} = μ_{0} {nAN}_{s} i_{0} ω ∴ f = \frac{ω}{2 π} = \frac{E_{\max}}{2 {πμ}_{0} {nAN}_{s} i_{0}} f = \frac{2.5 \times 10^{- 2}}{6.28 \times 12.56 \times 10^{- 7} \times 10^{5} \times 10^{- 4} \times 20 \times 1} = 15.85 s^{- 1}$

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A coil of area $7 {cm}^{2}$ and of 50 turns is kept with its plane normal to a magnetic field B. A resistance of 30 ohm is connected to the resistance-less coil. B is 75 exp (– 200t) gauss. The current passing through the resistance at t = 5 ms will be-

A.

$0.64 mA$

B.

$1.05 mA$

C.

$1.75 mA$

D.

$14 2.60 mA$

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Explanation

$1 . E = - NA \frac{dB}{dt} ∴ i = \frac{E}{R} = \frac{NA}{R} \frac{d}{dt} (75 e^{- 200 t}) \times 10^{- 4} = \frac{NA}{R} [75 \times (- 200) e^{- 200 t}] \times 10^{- 4} = + \frac{50 \times 7 \times 10^{- 8}}{30} [15000 e^{- 1}] = \frac{175 \times 10^{- 5}}{e} = \frac{175 \times 10^{- 5}}{2.73} = 0.64 \times 10^{- 3} A i = 0.64 mA$

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