Magnetic Effects Of Current(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

MAGNETIC EFFECTS OF CURRENT MCQs

Total Questions : 30 | Page 1 of 3 pages

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 2.

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

$\frac{μ_{0} i}{r}$
$\frac{μ_{0} i}{2 r}$
$\frac{μ_{0} i}{4 r}$
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 AB when extended.

Question 3.

A current I flows along the length of an infinitely long, straight and thin-walled pipe. Choose the correct option?

The magnetic field at all points inside the pipe is the same but not zero
The magnetic field at any point inside the pipe is zero
The magnetic field is zero only on the axis of the pipe
The magnetic field is different at different points inside the pipe

Discuss Question

Answer: Option B. -> The magnetic field at any point inside the pipe is zero
:
B
Applying Ampere's law

\oint B . d l = μ_{0} i

to any closed path inside the pipe, we can understand that as the currect enclosed inside it is zero, the magnetic field will be equal to zero.

Question 4.

A long solenoid has 200 turns per cm and carries a current of 2.5 amps. The magnetic field at its centre is ( $μ_{0} = 4 π \times 10^{- 7} w e b e r / a m p - m$ )

$3.14 \times 10^{2} w e b e r / m^{2}$
$6.28 \times 10^{- 2} w e b e r / m^{2}$
$9.42 \times 10^{- 2} w e b e r / m^{2}$
$12.56 \times 10^{- 2} w e b e r / m^{2}$

Discuss Question

Answer: Option B. ->

6.28 \times 10^{- 2} w e b e r / m^{2}

:
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 5.

Two long wires are hanging freely. They are joined first in parallel and then in series and then are connected with a battery. In both cases, which type of force acts between the two wires

Attraction force when in parallel and repulsion force when in series
Repulsion force when in parallel and attraction force when in series
Repulsion force in both cases
Attraction force in both cases

Discuss Question

Answer: Option A. -> Attraction force when in parallel and repulsion force when in series
:
A
When connected in parallel the current will be in the same direction hence the wires attract each other and when connected in series the current will be in the opposite direction. Hence they repel each other.

Two Long Wires Are Hanging Freely. They Are Joined First In ...

Question 6.

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

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

Discuss Question

Answer: Option D. ->

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

Current I Flows Through A Long Conducting Wire Bent At Right...

Question 7.

Two very long, straight and parallel wires carry steady currents I and I respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

$\frac{μ_{0} I q v}{2 π d}$
$\frac{μ_{0} I q v}{π d}$
$\frac{2 μ_{0} I q v}{π d}$
$0$

Discuss Question

Answer: Option D. ->

0

:
D
According to gives information following figure can be drawn, which shows that direction of magnetic field is along the direction of motion of charge so net force on it is zero.
if the currents are parallel then the net magnetic field at midpoint will be zero, Hence, no magnetic force acts on it.

Two Very Long, Straight And Parallel Wires Carry Steady Curr...

Question 8.

A long vertical wire carrying a current of 1A in the upward direction is placed in a region where a horizontal magnetic field of $2.0 \times 10^{- 4} T$ exists from south to north. Find the point where the resultant magnetic field is zero.

1.0mm, to the west of the wire
1.0 mm, to the East of the wire
2.0 mmm to the North of the wire
2.0 mm, to the South of the wire

Discuss Question

Answer: Option A. -> 1.0mm, to the west of the wire
:
A

B = B_{H} \frac{μ_{0} i}{2 π r} = B_{H} \frac{2 \times 10^{- 7} \times 1}{r} = 2.0 \times 10^{- 4}

r = 10^{- 3} m

west of the wire
= 1.0 mm west of the wire

A Long Vertical Wire Carrying A Current Of 1A In The Upward ...

Question 9.

The magnetic field at the centre of the circular coil carrying current of 4A is

$\frac{8 π}{3} \times 10^{- 5} T$
$\frac{8 π}{3} \times 10^{- 4} T$
$2 π \times 10^{- 5} T$
$2 π \times 10^{- 4} T$

Discuss Question

Answer: Option C. ->

2 π \times 10^{- 5} T

:
C

B = \frac{μ_{0} i}{2 r} [\frac{θ}{2 π}] = \frac{4 π 10^{- 7} \times 4}{2 \times 3 \times 10^{- 2}} [\frac{\frac{3 π}{2}}{2 π}] = 2 π \times 10^{- 5} T

Question 10.

A closed circuit is in the form of a regular hexagon of side a. If the circuit carries current I, What is magnetic induction at the cetre of the hexagon?