Magnetic Effects Of Current(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

MAGNETIC EFFECTS OF CURRENT MCQs

Magnetic Effects Of Electric Current, Magnetic Effects Of Current(10th And 12th Grade)

Total Questions : 54 | Page 3 of 6 pages

Question 21. Current i flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at a distance r from O is

μ0iπr
2μ0iπr
μ0i4πr(√2+1)
μ04π.2ir(√2+1)

Discuss Question

Answer: Option D. -> μ04π.2ir(√2+1)
:
D
By using

B = \frac{μ_{0}}{4 π} . \frac{i}{d} (s i n ϕ_{1} + s i n ϕ_{2})

, from figure

d = r s i n 45^{\circ} = \frac{r}{\sqrt{2}}

Magnetic field due to each wire at P

B = \frac{μ_{0}}{4 π} . \frac{i}{(r / \sqrt{2})} (s i n 45^{\circ} + s i n 90^{\circ})

= \frac{μ_{0}}{4 π} . \frac{i}{r} (\sqrt{2} + 1)

Hence net magnetic field at P

B_{n e t} = 2 \times \frac{μ_{0}}{4 π} . \frac{i}{r} (\sqrt{2} + 1)

Question 22. In the figure, shown the magnetic induction at the centre of there arc due to the current in portion AB will be

μ0ir
μ0i2r
μ0i4r
Zero

Discuss Question

Answer: Option D. -> Zero
:
D
The magnetic induction at O due to the current in portion AB will be zero because it lies on ABwhen extended.

Question 23. A straight wire of length

(π^{2})

metre is carrying a current of 2A and the magnetic field due to it is measured at a point distant 1 cm from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be

50 : 1
1 : 50
100 : 1
1 : 100

Discuss Question

Answer: Option B. -> 1 : 50
:
B
If a wire of length l is bant in the form of a circle of radius r then

2 π r = l \Rightarrow r = \frac{l}{2 π}

⃗ F = i (⃗ L \times ⃗ B)

Magnetic field due to straight wire

B_{1} = \frac{μ_{0}}{4 π} . \frac{2 i}{r} = \frac{μ_{0}}{4 π} \times \frac{2 \times 2}{1 \times 10^{- 2}}

also magnetic
field due to circular loop

B_{2} = \frac{μ_{0}}{4 π} . \frac{2 π i}{r} = \frac{μ_{0}}{4 π} . \frac{2 π \times 2}{π / 2} \Rightarrow \frac{B_{2}}{B_{1}} = \frac{1}{50}

A Straight Wire Of Length (π2) metre Is Carrying A Current...

Question 24. A long solenoid has 200 turns per cm and carries a current of 2.5 amps. The magnetic field at its centre is (

μ_{0} = 4 π \times 10^{- 7} w e b e r / a m p - m

)

3.14×102 weber/m2
6.28×10−2 weber/m2
9.42×10−2 weber/m2
12.56×10−2 weber/m2

Discuss Question

Answer: Option B. -> 6.28×10−2 weber/m2
:
B

B = μ_{0} n i = 4 π \times 10^{- 7} \times \frac{200}{10^{- 2}} \times 2.5 = 6.28 \times 10^{- 2} W b / m^{2}

Question 25. A current i is flowing in a straight conductor of length L. The magnetic field at a point distant

\frac{L}{4}

from its centre will be

4μ0i√5πL
4μ0i2πL
μ0i√2L
Zero

Discuss Question

Answer: Option A. -> 4μ0i√5πL
:
A

A Current I Is Flowing In A Straight Conductor Of Length L. ...

By using

B = \frac{μ_{0}}{4 π} . \frac{i}{a} (s i n ϕ_{1} + s i n ϕ_{2})

\Rightarrow B = \frac{μ_{0}}{4 π} . \frac{i}{(L / 4)} (2 s i n ϕ)

Also

s i n ϕ = \frac{L / 2}{\sqrt{5} L / 4} = \frac{2}{\sqrt{5}}

\Rightarrow B = \frac{4 μ_{0} i}{\sqrt{5} π L}

Question 26. The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

π24√2
π28√2
π2√2
π4√2

Discuss Question

Answer: Option B. -> π28√2
:
B

The Ratio Of The Magnetic Field At The Centre Of A Current C...

Magnetic field at the centre of circular coil

B_{c i r c u l a r} = \frac{μ_{0}}{4 π} . \frac{2 π i}{r} = \frac{μ_{0}}{4 π} . \frac{4 π^{2} i}{L}

Magnetic field at the centre of square coil

B_{s q u a r e} = \frac{μ_{0}}{4 π} . \frac{8 \sqrt{2} i}{a} = \frac{μ_{0}}{4 π} . \frac{32 \sqrt{2} i}{L}

Hence

\frac{B_{c i r c u l a r}}{B_{s q u a r e}} = \frac{π^{2}}{8 \sqrt{2}}

Question 27. Two infinitely long insulated wires are kept perpendicular to each other. They carry currents

I_{1} = 2 A

and

I_{2} = 1.5 A

.
i) Find the magnitude and direction of the magnetic field at P.
ii) If the direction of current is reversed in one of the wires. What would be the magnitude of the field B?

Two Infinitely Long Insulated Wires Are Kept Perpendicular T...

2×10−5 T, normally into the plane of the paper, 2√2×10−5 T
2×10−5 T,normally out of the plane of the paper, 2√2×10−5 T
2×10−5 T, normally into the plane of the paper, Zero
2×10−5 T, normally into the plane of the paper, Zero

Discuss Question

Answer: Option C. -> 2×10−5 T, normally into the plane of the paper, Zero
:
C

B_{1} = \frac{μ_{0} i_{1}}{2 π r_{1}} = \frac{2 \times 10^{- 7} \times 2}{4 \times 10^{- 2}} = 10^{- 5} B_{2} = \frac{μ_{0} i_{2}}{2 π r_{2}} = \frac{2 \times 10^{- 7} \times 1.5}{3 \times 10^{- 2}} = 10^{- 5} B = B_{1} + B_{2} = 10^{- 5} + 10^{- 5}

= 2 \times 10^{- 5}

, normally in to the place of the paper, zero

Question 28. Two thin long parallel wires separated by a distance b are carrying a current i amp each. The magnitude of the force per unit length exerted by one wire on the other is

μ0i2b2
μ0i22πb
μ0i2πb
μ0i2πb2

Discuss Question

Answer: Option B. -> μ0i22πb
:
B
Force per unit length

= \frac{μ_{0}}{4 π} . \frac{2 i_{1} i_{2}}{r} = \frac{μ_{0}}{2 π} . \frac{i^{2}}{b}

Question 29. Field inside a solenoid is

Directly proportional to its length
Directly proportional to current
Inversely proportional to total number of turns
Inversely proportional to current

Discuss Question

Answer: Option B. -> Directly proportional to current
:
B

B = μ_{0} n i

Question 30. A helium nucleus makes a full rotation in a circle of radius 0.8 metre in two seconds. The value of the magnetic field B at the centre of the circle will be
[CPMT 1988; KCET 1998; UPSEAT 2001]

Discuss Question

Answer: Option B. -> Directly proportional to current
:
B