Magnetic Effects Of Current Test

Result

Magnetic Effects Of Current Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

A particle consisting of two electrons is moving in a magnetic field of (3i+2j)T a velocity The magnetic force acting on the particle will be

A

B

C
zero

D

Solution
Question 2
1 / -0

What is the net force on the square coil

A
25×10⁻⁷ N moving towards wire

B
25×10⁻⁷ N moving away wire

C
35×10⁻⁷ N moving towards wire

D
35×10⁻⁷ N moving away wire

Solution

Force on side BC and AD are equal but opposite so their net will be zero.
Question 3
1 / -0

A proton is to circulate the earth along the equator with a speed of 1⋅0×107ms⁻¹. Find the minimum magnetic field which should be created at the equator for this purpose. The mass of proton = 1⋅7×10⁻²⁷kg and radius of earth = 6⋅37×10⁶m

A
1.6 × 10^–19

B
1.67 × 10^–8

C
1.0 × 10^–7

D
2 × 10^–7

Solution
Question 4
1 / -0

An α-particle is describing a circle of radius 0.45 m in a field of magnetic induction 1.2 weber/m². The potential difference required to accelerate the particle, so as to give this much energy to it (The mass of α-particle is 6.8 × 10^-27kg and its charge is 3.2 × 10^–19coulomb.) will be-

A
6 × 10⁶ V

B
2.3 × 10^–12V

C
7 × 10⁶ V

D
3.2 × 10^–12 V

Solution

If V is accelerating potential of α-particle, then Kinetic energy = qV 14 × 10⁶ eVolt = 2eV (since charge on α-particle = 2e)
Question 5
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An electron beam passes through a magnetic field of 2×10⁻³ weber/m² and an electric field of 1⋅0×10⁴Vm⁻¹ both acting simultaneously. The path of electrons remaining undeviated, calculate the speed of the electrons. If the electric field is removed what will be the radius of the electron path?

A
1.43 × 10^-2 m

B
0.43 × 10^-2 m

C
2.43 × 10^-2 m

D
3.43 × 10^-2 m

Solution
Question 6
1 / -0

A one meter long wire is lying at right angles to the magnetic field. A force of 1 kg wt is acting on it in a magnetic field of 0.98 Tesla. The current flowing in it will be

A
100 A

B
10 A

C
1 A

D
Zero

Solution

F = Bil ⇒ 1 × 9.8 = 0.98 × i × 1 ⇒ i = 10A
Question 7
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A circular coil of radius 4 cm and 20 turns carries a current of 3 ampere. It is placed in a magnetic field of 0.5 T. The magnetic dipole moment of the coil is

A
0.60 A-m²

B
0.45 A-m²

C
0.3 A-m²

D
0.15 A-m²

Solution
Question 8
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A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of 0.1 weber/m². The amount of work done in rotating it through 180° from its equilibrium position will be

A
0.1 J

B
0.2 J

C
0.4

D
0.8 J

Solution

Work done in rotating a coil through an angle θ from it's equilibrium position is W = MB(1 – cosθ) where θ = 180° and M

= W = 0.1 j
Question 9
1 / -0

The deflection in a moving coil galvanometer is

A
Directly proportional to the torsional constant

B
Directly proportional to the number of turns in the coil

C
Inversely proportional to the area of the coil

D
Inversely proportional to the current flowing

Solution

The deflection of the coil of moving coil galvanometer is given by:

⇒ θ ∝ N (Number of turns)
Question 10
1 / -0

A moving coil galvanometer has a coil of effective area A and number of turns N. The magnetic field B is radial. If a current I is passed through the coil, the torque acting on the coil is.

A
NA²B²I

B
NABI²

C
N²ABI

D
NABI

Solution

Let l and b be the length and breadth of the coil respectively. Effective area =A

If θ is the angle formed by direction of magnetic field with normal drawn on the plane of coil , then

Torque τ = NIBA sin θ

With radial magnetic field θ = 90

τ = NABI