Test: Ampere’s Circuital Law (December 29) - NEET MCQ

# Test: Ampere’s Circuital Law (December 29) - NEET MCQ

Test Description

## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Ampere’s Circuital Law (December 29)

Test: Ampere’s Circuital Law (December 29) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Ampere’s Circuital Law (December 29) questions and answers have been prepared according to the NEET exam syllabus.The Test: Ampere’s Circuital Law (December 29) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Ampere’s Circuital Law (December 29) below.
Solutions of Test: Ampere’s Circuital Law (December 29) questions in English are available as part of our Daily Test for NEET Preparation for NEET & Test: Ampere’s Circuital Law (December 29) solutions in Hindi for Daily Test for NEET Preparation course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Ampere’s Circuital Law (December 29) | 10 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study Daily Test for NEET Preparation for NEET Exam | Download free PDF with solutions
Test: Ampere’s Circuital Law (December 29) - Question 1

### Ampere's circuital law is given by

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 1

The line integral of the magnetic field of induction  around any closed path in free space is equal to absolute permeability of free space μ0 times the total current flowing through area bounded by the path.
Ampere's circuital law is given by:

Test: Ampere’s Circuital Law (December 29) - Question 2

### Two identical current carrying coaxial loops, carry current I in opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C, then which statement is correct?

 1 Crore+ students have signed up on EduRev. Have you?
Test: Ampere’s Circuital Law (December 29) - Question 3

### A long straight wire in the horizontal plane carries a current of 75 A in north to south direction, magnitude and direction of field B at a point 3 m east of the wire is

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 3

From Ampere circuital law

The direction of field at the given point will be vertical up determined by the screw rule or right hand rule.

Test: Ampere’s Circuital Law (December 29) - Question 4

If a long straight wire carries a current of 40 A, then the magnitude ol the field B at a point 15 cm away from the wire is

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 4

I = 40A
r = 15 cm = 15 x 10-2 m
∴  = 5.34 x 10-5 T

Test: Ampere’s Circuital Law (December 29) - Question 5

The correct plot of the magnitude of magnetic field   vs distance r from centre of the wire is, if the radius of wire is R

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 5

The magnetic field from the centre of wire of radius R is given by
B = ((μ0I)/(2R2))r (r < R) ⇒ B ∝ r
and B = μ0I/2πr (r > R) ⇒ B ∝ 1/r
From this descriptions, we can say that the graph (b) is a correct representation.

Test: Ampere’s Circuital Law (December 29) - Question 6

A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns. If it carries a current of 4A, then the magnitude of the magnetic field inside the solenoid is

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 6

Here, n = 600/0.6 = 1000 turns/m, I = 4A
l = 0.6m, r = 0.02 m ∵ 1/r = 30 i.e., l >>r
Hence, we can use long solenoid formuls, then
∴ B = μ0nI = 4π x 10-7 x 103 x 4 = 50.24 x 10-4
= 5.024 x 10-3 T

Test: Ampere’s Circuital Law (December 29) - Question 7

A solenoid of length 50 cm, having 100 turns carries a current of 2.5 A. The magnetic field at one end of the solenoid is

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 7

Here, I = 2.5 A , I = 50 cm = 0.50 m and n = 100/0.50 = 200 m-1
∴

Test: Ampere’s Circuital Law (December 29) - Question 8

A 90 cm long solenoid has six layers of windings of 450 turns each. If the diameter of solenoid is 2.2 cm and current carried is 6A, then the magnitude of magnetic field inside the solenoid, near its centre is

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 8

For six layers of windings the total number of turns = 6 x 450 = 2700
Now number of turns per unit length

Then the field in side the solenoid near the centre
B = μ0nI = 4π x 10-7 x 3000 x 6  = 72π x 10-4 T = 72πG

Test: Ampere’s Circuital Law (December 29) - Question 9

Which of the following statement is incorrect?

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 9

Option C is incorrect as magnetic field inside the core of the toroid varies greatly.

Test: Ampere’s Circuital Law (December 29) - Question 10

The inner and outer radius of a toroid core are 28 cm and 29 cm respectively and around the core 3700 turns of a wire are wounded. If the current in the wire is 10A, then the magnetic field inside the core of the toroid is

Detailed Solution for Test: Ampere’s Circuital Law (December 29) - Question 10

The number of turns per unit length for the given toroid n = N/2πrav
The average radius of toroid
rav = (28 + 29)/2 = 28.5 cm 28.5 x 10-2 m

## Daily Test for NEET Preparation

12 docs|366 tests
Information about Test: Ampere’s Circuital Law (December 29) Page
In this test you can find the Exam questions for Test: Ampere’s Circuital Law (December 29) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Ampere’s Circuital Law (December 29), EduRev gives you an ample number of Online tests for practice

## Daily Test for NEET Preparation

12 docs|366 tests