Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - NEET MCQ

# Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - NEET MCQ

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## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27)

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Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 1

### If we double the radius of a current carrying coil keeping the current unchanged. what happens to the magnetic field at its Centre?

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 1

As,
B=μonI/2a
B ∝1/a
B1/B2=a2/a1
B1=2B2
B2=(1/2) x B1,
Magnetic field is halved.

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 2

### A circular arc of metallic wire subtends an angle π/2 at center. If it carries current I and its radius of curvature is r, then the magnetic field at the center of the arc is

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 2

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Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 3

### The magnetic field B on the axis of a circular coil at distance x far away from its centre are related as:

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 3

dB=μoidlsin90/4πr2oidl/4πr2
Bnet= ∫dBsinθ
= ∫μoidlR/4πr2r== (μoiR/4πr3) ∫dl
= (μoiR/4πr3)2πR= μoiR2/2(R2+x2)3/2
B= μoiR2/2(R2+x2)3/2
If x>>>R
B= μoiR2/2(x2)3/2

[=B ∝ x-3]

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 4

Two concentric coils carry the same current in opposite directions. The diameter of the inner coil is half that of the outer coil. If the magnetic field produced by the outer coil at the common centre are 1 T, the net field at the centre is

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 4

The magnetic field produced by a current-carrying coil at its center is given by the formula,
B = μ0 * (N*I/R),
where,
B is the magnetic field,
μ0 is the permeability of free space,
N is the number of turns in the coil,
I is the current through the coil, and
R is the radius of the coil.
In this case, both the coils carry the same current but in opposite directions. So, the fields produced by them will be in opposite directions. Also, the diameter of the inner coil is half that of the outer coil. Thus, the radius of the inner coil will be half that of the outer coil.
Therefore, the field at the center due to the inner coil will be double that due to the outer coil (because the magnetic field is inversely proportional to the radius).
Since the fields are in opposite directions, the net field at the center will be the difference between the two fields. That is, 2B (due to the inner coil) - B (due to the outer coil) = B.
So, if the field due to the outer coil is 1 T (Tesla), the net field at the center will also be 1 T.
Hence, the correct answer is 3. 1T.

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 5

The magnetic field due to current element depends upon which of the following factors:

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 5

From Biot-Savart law, magnetic field at a point p, B= (μ0​/4π)∫ [(Idl×r​)/ r3]
where r is the distance of point p from conductor and I is the current in the conductor.
Thus magnetic field due to current carrying conductor depends on the current flowing through conductor and distance from the conductor and length of the conductor.

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 6

A circular coil of radius r carries current I. The magnetic field at its center is B. at what distance from the center on the axis of the coil magnetic field will be B/8

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 6

B= 2μ0​i⋅πr2/4π(r2+x2)3/2
Magnetic field at centre= μ0​i/2r
B= μ0​i/2r
Put r= √3
B1=2μ0​i⋅πr2/4π(4r2)3/2
=B/8
Hence, √3R distance from the centre magnetic field is equal to magnetic field at centre

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 7

A circular loop of radius 0.0157 m carries a current of 2 A. The magnetic field at the centre of the loop is​

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 7

The magnetic field due to a circular loop is given by:
B= μ0​​2πi​/4πr
=10−7×2π×2/0.0157 ​
=8×10−5 Wb/m2

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 8

A straight conductor carrying current I is split into circular loop as shown in figure a , the magnetic induction at the center of the circular loop is

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 8

The magnetic field at the center O due to the upper side of the semicircular current loop is equal and opposite to that due to the lower side of the loop.

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 9

The magnetic field due to circular coil of 200 turns of diameter 0.1m carrying a current of 5A at a point on the axis of the coil at a distance 0.15m from the center of the coil will be

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 9

B=μ02πnIa2/4π (a2+x2)3/2
=10−7×2×(22/7)×200×5×(0⋅1/2)2/ [(0⋅1/2)2+(0⋅15)2]3/2
=39.74x10-5

Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 10

Wire of length l, carries a steady current I. It is bent first to form a circular coil of one turn. The same wire of same length is now bent more sharply to give two loops of smaller radius the magnetic field at the centre caused by the same current is

Detailed Solution for Test: Magnetic Field due to a Current Element Biot-Savart Law (December 27) - Question 10

Let the radii be r1​ and r2​ respectively.
Since there are two turns of radius r2​, r1​=2r2​
Magnetic field B at the centre of  the coil of radius r1​ B1​=​μo​i/2r1​=​μo​i​/4r2
Magnetic field B at the center of the coil of radius r2​ B2​=2×​μo​i​/2r2
∴ B2/B1 =(2× μo​i/2r2​)/(μo​i /4r2​)​ ​​=4
Hence the answer is option C, four times its initial value.

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