Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 17

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Question 1
1 / -0

If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be

A
Only inside the pipe

B
Only outside the pipe

C
Neither inside nor outside

D
Both inside and outside the pipe

Solution

According to Ampere's circuital law,

B(2πr) = μ0 × 0

Or B = 0

So, inside a hollow metallic (copper) pipe carrying current, the magnetic field is zero.

But for external points, the whole current behaves as if were concentrated at the axis only, so outside

B0 = μ0i/2πr

Then, the magnetic field is produced outside the pipe only.
Question 2
1 / -0

The number of lines of force passing through a unit area placed perpendicularly to the magnetic lines of forces is termed as

A
Magnetic induction

B
Magnetic flux intensity

C
Intensity of magnetic field

D
All of above
Question 3
1 / -0

Magnetic effect of current was discovered by:

A
Faraday

B
Oersted

C
Ampere

D
Bohr

Solution

The magnetic effect of electric current was discovered by H.C. Oersted in 1820. He observed that flow of electric current through a conductor produces a magnetic field around it.
Question 4
1 / -0

Assertion: Cyclotron does not accelerate electron.

Reason: Mass of the electron is very small.

(A) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(B) If both assertion and reason are true but reason is not the correct explanation of the assertion.

(C) If assertion is true but reason is false.

(D) If the assertion and reason both are false.

A
A

B
B

C
C

D
D

Solution

Cyclotron is suitable for accelerating heavy particles like protons, a-particles, etc, and not for electrons because of low mass. Because electrons acquire very high velocities very near to velocity of light and appreciable variation in their mass, occurs.
Question 5
1 / -0

A one metre 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