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 Multiple Choice QuestionsMultiple Choice Questions

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1.

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 µT. What will be its value at the centre of the loop?

  • 250 µT

  • 150 µT

  • 125 µT

  • 125 µT


A.

250 µT

Using space formula space straight B space equals fraction numerator straight mu subscript 0 space iR squared over denominator 2 left parenthesis straight R squared space plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction comma
we space get
54 space equals fraction numerator straight mu subscript 0 space straight i space left parenthesis 3 right parenthesis squared over denominator 2 left parenthesis straight R squared space plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction
At space the space centre space of space the space coil comma space straight X equals 0 space and space straight B equals fraction numerator straight mu subscript 0 space straight i over denominator 2 left parenthesis 3 right parenthesis end fraction
using space equation space left parenthesis straight i right parenthesis
straight B space equals fraction numerator 54 space straight x space 5 cubed over denominator left parenthesis 3 right parenthesis squared space straight x space 3 end fraction
equals straight B space equals space 250 space μT
625 Views

2.

A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is

  • fraction numerator 2 πmq over denominator straight B end fraction
  • fraction numerator 2 πq squared straight B over denominator straight m end fraction
  • fraction numerator 2 πqB over denominator straight m end fraction
  • fraction numerator 2 πqB over denominator straight m end fraction

D.

fraction numerator 2 πqB over denominator straight m end fraction
mω squared straight r space equals Bqωr
straight omega space equals space Bq divided by straight m
straight T space equals fraction numerator 2 πm over denominator qB end fraction
158 Views

3.

In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

  • 200Ω

  • 100Ω

  • 500 Ω

  • 500 Ω


B.

100Ω

The potential drop across the resistance R is 2V. Now, no current flows through the galvanometer .and if V1 is the potential drop across the resistor R and V is the total potential then
fraction numerator VR over denominator 500 space plus straight R space end fraction space space equals space straight V subscript 1

fraction numerator 12 space straight R over denominator 500 space plus straight R space end fraction space equals space 2
space equals space 100 space ohm

422 Views

4.

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be

  • nB

  • n2B

  • 2nB

  • 2nB


B.

n2B

The magnetic field at the centre of circular coil is

straight B space equals fraction numerator straight mu subscript 0 space straight i over denominator 2 straight r end fraction
where space straight r space equals space radius space of space circle space space equals space fraction numerator 1 over denominator 2 straight pi end fraction space left parenthesis because space straight I space equals 2 πr right parenthesis
straight B space equals space fraction numerator straight mu subscript 0 space straight i space over denominator 2 end fraction space straight x space fraction numerator 2 straight pi over denominator straight I end fraction
space equals space fraction numerator straight mu subscript 0 space straight i space straight pi over denominator straight I end fraction space.... space left parenthesis straight i right parenthesis

When wire of length i bents into a circular loops of n turns, then

l = n × 2 π r

⇒ r = 1/ n x 2 π

Thus, new magnetic field

straight B apostrophe space equals space fraction numerator straight mu subscript 0 ni over denominator 2 straight r apostrophe end fraction space equals space fraction numerator straight mu subscript 0 ni over denominator 2 end fraction space straight x space fraction numerator straight n space straight x space 2 straight pi over denominator straight l end fraction
space equals space fraction numerator straight mu subscript 0 space straight i space straight pi over denominator straight l end fraction space straight x space straight n squared
space equals space straight n squared straight B

1355 Views

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5.

One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. if each tube moves towards the other at a constant speed V, then the emf induced in the circuit in terms of B,

  • B

  • -B

  • zero

  • zero


D.

zero

open vertical bar dϕ over dt close vertical bar space equals space 2 straight B calligraphic l v
325 Views

6.

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be

  • 103

  • 105

  • 99995

  • 99995


D.

99995

straight I subscript straight g space equals space 15 space mA
straight V subscript straight g space equals space 75 space mV
straight R space equals space straight V over straight I subscript straight g minus straight V subscript straight g over straight I subscript straight g
210 Views

7.

A current I ampere flows along an infinitely long straight thin-walled tube, then the magnetic induction at any point inside the tube is

  • infinite

  • zero

  • fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator 2 straight i over denominator straight r end fraction space tesla
  • fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator 2 straight i over denominator straight r end fraction space tesla

B.

zero

Let R be the radius of a long thin cylindrical shell. To calculate the magnetic induction at a distance r (r < R) from the axis of cylinder, a circular shell of radius r is shown:

Since no current is enclosed in the circle so, from Ampere's circuital law, magnetic induction is zero at every point of the circle. Hence, the magnetic induction at any point inside the infinitely long straight thin-walled tube (cylindrical) is zero.

452 Views

8.

A charged oil drop is suspended in a uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10−15 kg and g = 10 m/s2 )

  • 3.3 × 10−18 C

  • 3.2 × 10−18 C

  • 1.6 × 10−18 C

  • 1.6 × 10−18 C


A.

3.3 × 10−18 C

In steady state, electric force on drop = weight of drop
∴ qE = mg



straight q space equals space mg over straight E
space equals space fraction numerator 9.9 space space straight x space 10 to the power of negative 15 end exponent space straight x space 10 over denominator 3 space straight x space 10 to the power of 4 end fraction space equals space 3.3 space straight x space 10 to the power of negative 18 end exponent space straight C

155 Views

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9.

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then

  • its velocity will decrease.

  • its velocity will increase.

  • it will turn towards right of direction of motion.

  • it will turn towards right of direction of motion.


A.

its velocity will decrease.

straight F with rightwards arrow on top space equals space minus straight e space open square brackets straight E with rightwards arrow on top space plus space straight v with rightwards arrow on top straight x space straight B with rightwards arrow on top close square brackets space equals negative straight e straight E with rightwards arrow on top
straight a with rightwards arrow on top space equals space minus space fraction numerator straight e straight E with rightwards arrow on top over denominator straight m end fraction
straight v left parenthesis straight t right parenthesis space equals space straight v subscript straight o space minus eE over straight m straight t
135 Views

10.

In a uniform magnetic field of induction B a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R the mean power generated per period of rotation is

  • fraction numerator Bπr squared space straight omega over denominator 2 straight R end fraction
  • fraction numerator left parenthesis Bπr squared straight omega right parenthesis squared over denominator 2 straight R end fraction
  • fraction numerator left parenthesis Bπrω squared right parenthesis over denominator 2 straight R end fraction
  • fraction numerator left parenthesis Bπrω squared right parenthesis over denominator 2 straight R end fraction

B.

fraction numerator left parenthesis Bπr squared straight omega right parenthesis squared over denominator 2 straight R end fraction
Magnetic space flux space equals space BA space cos space straight theta space equals space straight B. πr squared over 2 space cos space ωt
therefore space straight P space equals straight epsilon squared subscript ind over straight R space equals space fraction numerator straight B squared straight pi squared straight r to the power of 4 straight omega squared space sin squared space ωt over denominator 4 straight R end fraction
Now comma space less than sin squared space ωt greater than space equals space 1 divided by 2 space left parenthesis mean space value right parenthesis
therefore space less than straight P greater than space equals space fraction numerator left parenthesis Bπr squared straight omega right parenthesis squared over denominator 8 straight R end fraction
571 Views

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