Magnetic Effects Of Current Test

Result

Magnetic Effects Of Current Test - 5

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Select correct statement(s) regarding cyclotron

A
Charged particle accelerated only between two dees

B
Charged particle accelerated throughout its circular path

C
Charged particle is accelerated in the dees but its speed up only between the two dees

D
Both (b) & (c)

Solution

Charge particles accelerate due to magnetic field in dees and due to electric field in gap.
Question 2
1 / -0

A short wire of length L and mass m is placed along northsouth direction on rough surface of friction coefficient . A charge q instantly flows through this wire. Take horizontal and vertical components of magnetic field of earth B_H and B_V respectively. The distance travelled by wire before it comes to rest.

A

B

C

D
Zero

Solution

Let instantaneous current in wire due to charge flows I = dq/dt

F = B_vIL (impulsive horizontal force due to current)

The wire gains velocity instantly will come to rest due to friction
Question 3
1 / -0

If Ab = Bc = L, Then The Net Magnetic Force On The Rod Is

A
B₀I₀ 2L

B
B₀I₀ L

C
ZERO

D

Solution

The net magnetic force is given
Question 4
1 / -0

A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field directed into the paper. AO = l and OC = 3l. Then

A

B

C

D

Solution
Question 5
1 / -0

A charged particle enters a uniform magnetic field with velocity vector making an angle of 30° with the magnetic field. The particle describes a helical trajectory of pitch x. The radius of the helix is

A
x/2π

B

C

D

Solution

From equation (1) and (2), we get r =
Question 6
1 / -0

A wire of length metres carrying a current I amperes is bent in the form of a circle. The magnitude of the magnetic moment is

A

B

C

D

Solution

Magnetic moment m = AI = πr²I , where r is the radius of the circular loop. Now, the circumference of the circle = length of the wire, i.e.,
Question 7
1 / -0

Equal current I flows in circular wire segments ACB and ADB of equal radius r as shown in figure. If θ = 60°, the magnetic field at centre O is

A

B

C

D
zero

Solution

Magnetic field due to a complete circular loop at its centre is

∴ Magnetic field due to segment ADB at centre O is

Magnetic field due to segment ACB at centre O is
Question 8
1 / -0

Find The Magnetic Field At Point P As Shown In Figure Below

A

B

C

D

Solution
Question 9
1 / -0

A charged particle moves through a magnetic field directed perpendicular to its direction of motion. Which of the following quantities of the particle will not change?

A
Momentum

B
Speed

C
Velocity

D
None of these

Solution

Since, is perpendicular to

Hence, magnitude of will not change, but its direction will continuously change. That’s speed will not change, but velocity and momentum will change.
Question 10
1 / -0

A long solenoid of cross-sectional radius R has a thin insulated wire ring tightly put on its winding. One half of the ring has the resistance 10 times that of the other half. The magnetic induction produced by the solenoid varies with time as B = bt, whose b is a constant. Find the magnitude of the electric field strength in the ring.

A
9/11 Rb

B
9/22 Rb

C
9 Rb

D
Rb

Solution

Both upper half and lower half will have same effective area of πR²/2. so charge in flux will be same and induced emf will have same value. But since the resistance is different due to which current must be different but ring is as a whole is closed circuit so electric field will be generated to make the current flow in both parts to be same.

ε + EπR − i10r = 0 ..... (i)

ε − EπR − ir = 0 ..... (i)