CUET UG Physics Booster Test 2-Current Elements
📌 Answers are locked once submitted — results and explanations appear at the end.
QUESTION 1 OF 20
Match List I with List II for Biot-Savart properties.
| List I | List II |
|---|---|
| 1. Vector source | a. Represents an infinitesimal current element |
| 2. Total magnetic field | b. Assumed condition for Biot-Savart calculation |
| 3. Steady current | c. I dl |
| 4. I dl | d. Vector sum of individual source fields |
QUESTION 2 OF 20
Identify the correct statements regarding the displacement vector r in the Biot-Savart law.
Statements:
1. It extends from the source element dl to the point P.
2. Its magnitude determines the inverse-square falloff.
3. It is always perpendicular to dl.
4. It is used in the cross product dl×r.
QUESTION 3 OF 20
If the current I is increased by a factor of 3 and the length dl is doubled, while the distance r is halved, what is the ratio of the new magnetic field dB^'to the original field dB?
QUESTION 4 OF 20
If a magnetic field dB at distance r is formed by an element I dl, the magnetic field at a distance xr will be proportional to:
QUESTION 5 OF 20
Incorrect statement about the cross product dl×r in the Biot-Savart law.
QUESTION 6 OF 20
When applying the Right Hand Screw Rule to find the direction of dB, if the rotation from dl to r is clockwise, the resulting magnetic field is directed:
QUESTION 7 OF 20
The theoretical relationship between the speed of light c, permeability μ_0, and permittivity ε_0 is given by
c=(xμ_0ε_0)^(-y)
The valid values of x and y are:
QUESTION 8 OF 20
Identify the correct statements about the medium in the Biot-Savart law.
Statements:
1. The field formulation adapts if placed in another medium.
2. μ_0 relates strictly to the electrostatic field.
3. μ_0 is the permeability of free space.
4. The constant μ_0/4πis used when the medium is vacuum.
A.2 and 4 are correct
QUESTION 9 OF 20
Identify the correct statements comparing electric and magnetic field sources.
Statements:
1. Electrostatic field is produced by a scalar source.
2. Magnetic field is produced by a vector source I dl.
3. Both sources are vector at the macroscopic level.
4. Electric charge acts as the scalar source.
QUESTION 10 OF 20
Electrostatic field vs Magnetic field dependence on distance:
QUESTION 11 OF 20
A circular loop of radius R=3 cm carries a current. If the magnetic field at the center is B_0, what is the magnetic field on the axis at x=4 cm?
QUESTION 12 OF 20
When calculating the magnetic field on the axis of a circular current loop, the components perpendicular to the axis cancel out because:
QUESTION 13 OF 20
For a circular loop, the field at the center is B. If the current is scaled by x and the radius by y, the new field is:
QUESTION 14 OF 20
Incorrect statement regarding a tightly wound coil of N turns.
B=μ_0NI/2R
QUESTION 15 OF 20
When evaluating a circular current loop, the upper side and lower side can respectively be thought of as a:
QUESTION 16 OF 20
Identify the correct statements regarding the right-hand thumb rule for circular loops.
Statements:
1. The thumb points in the direction of the electric field.
2. Fingers point in the direction of the current.
3. The thumb gives the direction of the magnetic field.
4. Curl the palm around the wire.
QUESTION 17 OF 20
Identify the correct statements about the straight segments in a bent wire setup evaluated at the center of the arc.
Statements:
1. dl and r are parallel.
2. Their contribution to the magnetic field is zero.
3. The cross product dl×r=0.
4. They contribute a field equal to μ_0I/4πR.
QUESTION 18 OF 20
Identify the correct statements regarding points where the magnetic field contribution from an element dl is identically zero.
Statements:
1. Along the line of the element where θ=0^∘.
2. Where the displacement vector r is perpendicular to dl.
3. Where sinθ=0.
4. Directly on the straight line extending from the element.
QUESTION 19 OF 20
In a semi-circular arc carrying a steady current, the contribution to the magnetic field at the center resembles a full circular loop because:
QUESTION 20 OF 20
A current element of length Δx=2 cm is at the origin carrying 5 A. The magnetic field at a distance of 1 m on the y-axis is:
μ_0/4π=10^(-7) T m A^(-1)
Test Complete!
Answer Review
1 Match List I with List II for Biot-Savart properties.
| List I | List II |
|---|---|
| 1. Vector source | a. Represents an infinitesimal current element |
| 2. Total magnetic field | b. Assumed condition for Biot-Savart calculation |
| 3. Steady current | c. I dl |
| 4. I dl | d. Vector sum of individual source fields |
�� I dl is the source term. �� Steady current is assumed. �� Fields add by superposition.
In the Biot-Savart law, the source of the magnetic field is the vector quantity I dl. Hence, Vector source → I dl. The total magnetic field at a point is obtained by adding the contributions of all current elements vectorially. Therefore, Total magnetic field → Vector sum of individual source fields. The Biot-Savart law is applicable for steady currents. Thus, Steady current → Assumed condition for Biot-Savart calculation. Finally, I dl itself represents an infinitesimal current element. Therefore: 1-c, 2-d, 3-b, 4-a Hence, Option A is correct.
- �� Option B → Incorrectly matches vector source.
- �� Option C → Multiple mismatches occur.
- �� Option D → Incorrect assignments for source and field.
NCERT Recall
- Application
- Recall the meaning of each quantity appearing in the Biot-Savart law.
- Final Logic
- Source → I dl; Total field → Vector sum.
Source Is I dl, Field Is Sum
2 Identify the correct statements regarding the displacement vector r in the Biot-Savart law.
Statements:
1. It extends from the source element dl to the point P.
2. Its magnitude determines the inverse-square falloff.
3. It is always perpendicular to dl.
4. It is used in the cross product dl×r.
�� r joins source and observation point. �� Distance affects field strength. �� r need not be perpendicular to dl.
In the Biot-Savart law, the displacement vector r extends from the current element to the observation point P. Its magnitude appears in the denominator: dB∝1/r^2 Hence, it determines the inverse-square dependence of the magnetic field. The vector form of the law is: dB=μ_0/4πI(dl×r)/r^3 Therefore, r is involved in the cross product. However, r is not always perpendicular to dl. The angle between them may have any value from 0^∘to 180^∘. Thus, statements 1, 2 and 4 are correct.
- �� Option A → Includes statement 3, which is incorrect.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 3 is incorrect.
Concept Application
- Application
- Interpret the geometrical meaning of the displacement vector.
- Final Logic
- r joins source to observation point and controls distance dependence.
r Means Source To Point
3 If the current I is increased by a factor of 3 and the length dl is doubled, while the distance r is halved, what is the ratio of the new magnetic field dB^'to the original field dB?
�� dB∝I dl/r^2 �� Substitute the changed quantities. �� Compare new and original fields.
The Biot-Savart law gives: dB∝I dl/r^2 New values: I^'=3Idl^'=2dlr^'=r/2 Therefore: dB^'∝(3I)(2dl)/(r/2)^2dB^'∝6I dl/r^2/4dB^'∝24I dl/r^2 Hence: dB^'/dB=24 Therefore, Option C is correct.
- �� Option A → Ignores distance effect.
- �� Option B → Includes only part of the scaling.
- �� Option D → Opposite trend from actual dependence.
Substitution
- Application
- Apply proportionality directly.
- Final Logic
- 3×2×4=24
Triple Times Double Times Four
4 If a magnetic field dB at distance r is formed by an element I dl, the magnetic field at a distance xr will be proportional to:
�� Biot-Savart law follows inverse-square dependence. �� Replace r by xr. �� Compare both expressions.
The Biot-Savart law contains the distance dependence: dB∝1/r^2 If the distance becomes: r^'=xr then: dB^'∝1/(xr)^2dB^'=1/x^2(1/r^2) Therefore: dB^'=x^(-2)dB Thus, increasing the distance by a factor x decreases the magnetic field by a factor x^2. Hence, Option B is correct.
- �� Option A → Opposite dependence.
- �� Option C → Incorrect linear dependence.
- �� Option D → Represents inverse-first-power dependence.
NCERT Recall
- Application
- Use the inverse-square term in the Biot-Savart law.
- Final Logic
- dB∝1/r^2⇒dB^'∝x^(-2)dB
Distance Doubles Field Becomes One-Fourth
5 Incorrect statement about the cross product dl×r in the Biot-Savart law.
�� Cross products follow the right-hand rule. �� Anticlockwise rotation gives outward direction. �� Clockwise rotation gives inward direction.
The magnitude of a cross product is: ∣dl×r∣=dl r sinθ The resulting vector is perpendicular to the plane containing dl and r. According to the right-hand screw rule, if the rotation from the first vector to the second vector is anticlockwise, the resulting vector points outward from the plane (towards the observer), not away from the observer. Therefore, the statement that an anticlockwise movement produces a vector directed away from you is incorrect. Hence, Option C is the incorrect statement.
- �� Option A → Correct magnitude formula.
- �� Option B → Correct property of a cross product.
- �� Option D → Right-hand screw rule correctly determines direction.
Concept Application
- Application
- Apply the right-hand screw rule to determine vector direction.
- Final Logic
- Anticlockwise ⇒ Out of Plane; Clockwise ⇒ Into Plane.
Anti-Clockwise Comes Out
6 When applying the Right Hand Screw Rule to find the direction of dB, if the rotation from dl to r is clockwise, the resulting magnetic field is directed:
�� Direction is obtained using the right-hand screw rule. �� Clockwise rotation gives inward direction. �� Anticlockwise rotation gives outward direction.
In the Biot-Savart law, the magnetic field direction is determined by the cross product: dB∝dl×r The right-hand screw rule is used to determine the sense of the resulting vector. If the rotation from dl to r is clockwise, the screw advances away from the observer. Hence, the magnetic field vector points into the plane of the page. This direction is conventionally represented by a cross (×). If the rotation were anticlockwise, the field would point outward toward the observer and would be represented by a dot (∙). Therefore, clockwise rotation produces a magnetic field directed away from you. Hence, Option B is correct.
- �� Option A → Corresponds to anticlockwise rotation.
- �� Option C → Magnetic field is perpendicular to dl.
- �� Option D → Magnetic field is perpendicular to r.
Concept Application
- Application
- Apply the right-hand screw rule to the cross product dl×r.
- Final Logic
- Clockwise rotation ⇒ Into the page ⇒ Away from you.
Clockwise Goes Inward
7 The theoretical relationship between the speed of light c, permeability μ_0, and permittivity ε_0 is given by
c=(xμ_0ε_0)^(-y)
The valid values of x and y are:
�� Maxwell related light to electromagnetic fields. �� c depends on μ_0 and ε_0. �� The exponent is one-half.
Maxwell's electromagnetic theory predicts that electromagnetic waves travel with speed: c=1/√(μ_0ε_0) This can be rewritten as: c=(μ_0ε_0)^(-1/2) Comparing with: c=(xμ_0ε_0)^(-y) we obtain: x=1 and y=1/2 This equation provided one of the strongest theoretical arguments that light itself is an electromagnetic wave. Hence, Option A is correct.
- �� Option B → Incorrect coefficient and exponent.
- �� Option C → Exponent should be 1/2, not 2.
- �� Option D → Incorrect coefficient.
NCERT Recall
- Application
- Recall Maxwell's expression for the speed of light.
- Final Logic
- c=1/√(μ_0ε_0)
Light Speed Equals One By Root Mu Epsilon
8 Identify the correct statements about the medium in the Biot-Savart law.
Statements:
1. The field formulation adapts if placed in another medium.
2. μ_0 relates strictly to the electrostatic field.
3. μ_0 is the permeability of free space.
4. The constant μ_0/4πis used when the medium is vacuum.
A.2 and 4 are correct
�� μ_0 is a magnetic constant. �� Vacuum uses permeability of free space. �� Other media require modified permeability.
The permeability of free space is denoted by: μ_0 and appears in the Biot-Savart law: dB=μ_0/4πIdlsinθ/r^2 This form is valid when the medium is vacuum (free space). When magnetic fields exist in another medium, the magnetic permeability of that medium replaces the permeability of free space. Therefore, the field formulation adapts according to the properties of the medium. Statement 2 is incorrect because μ_0 is associated with magnetic fields, whereas electrostatics primarily involves the permittivity of free space ε_0. Hence: • Statement 1 is correct. • Statement 3 is correct. • Statement 4 is correct. • Statement 2 is incorrect. Therefore, the correct answer is C. 1, 3 and 4 are correct.
- �� Option A → Statement 2 is incorrect and Statement 3 is omitted.
- �� Option B → Statement 2 is incorrect because μ_0 is not an electrostatic constant.
- �� Option D → Statement 4 is also correct and should not be omitted.
NCERT Recall
- Application
- Recall the physical meaning of μ_0 and its role in the Biot-Savart law.
- Final Logic
- μ_0=Permeability of Free Space
- Vacuum uses μ_0, while other media use their own permeability.
Mu Means Magnetism
9 Identify the correct statements comparing electric and magnetic field sources.
Statements:
1. Electrostatic field is produced by a scalar source.
2. Magnetic field is produced by a vector source I dl.
3. Both sources are vector at the macroscopic level.
4. Electric charge acts as the scalar source.
�� Electric field originates from charge. �� Magnetic field originates from current elements. �� Electric charge is scalar.
Electrostatic fields arise from electric charge q, which is a scalar quantity. Hence: • Statement 1 is correct. • Statement 4 is correct. Magnetic fields arise from current elements represented by: I dl which possess both magnitude and direction. Therefore, the magnetic source is vector in nature. Thus, statement 2 is correct. Statement 3 is incorrect because electric charge remains a scalar quantity even at the macroscopic level. Therefore, statements 1, 2 and 4 are correct.
- �� Option A → Statement 3 is incorrect.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 2 is omitted.
Concept Application
- Application
- Differentiate between charge and current element.
- Final Logic
- Charge → Scalar Source; I dl→ Vector Source.
Charge Makes E Current Makes B
10 Electrostatic field vs Magnetic field dependence on distance:
�� Both laws contain 1/r^2. �� Both are long-range interactions. �� Both obey superposition.
Coulomb's law for the electric field due to a point charge is: E=1/4πε_0q/r^2 Thus: E∝1/r^2 Similarly, the Biot-Savart law for a current element is: dB=μ_0/4πIdlsinθ/r^2 Thus: dB∝1/r^2 Therefore, both electric and magnetic fields exhibit inverse-square distance dependence in their fundamental source laws. Hence, Option A is correct.
- �� Option B → Electrostatic field does not vary as 1/r^3.
- �� Option C → Magnetic field in Biot-Savart law does not vary as 1/r^3.
- �� Option D → Electrostatic field is not linear in distance.
NCERT Recall
- Application
- Compare the distance terms in Coulomb's and Biot-Savart laws.
- Final Logic
- E∝1/r^2,dB∝1/r^2
Both Follow Inverse Square Law
11 A circular loop of radius R=3 cm carries a current. If the magnetic field at the center is B_0, what is the magnetic field on the axis at x=4 cm?
�� Use the axial field formula. �� Compare with the center field. �� Express the answer in terms of B_0.
The magnetic field on the axis of a circular loop is: B=μ_0IR^2/2(R^2+x^2)^(3/2) At the center (x, 0): B_0=μ_0I/2R Therefore, B/B_0=R^3/(R^2+x^2)^(3/2) Given: R=3 cmx=4 cmR^2+x^2=3^2+4^2=25(R^2+x^2)^(3/2)=25^(3/2)=125R^3=27 Thus, B/B_0=27/125=0.216 Hence, B=0.216B_0 Therefore, Option A is correct.
- �� Option A → Uses (R/x)^3 incorrectly.
- �� Option B → Incorrect ratio calculation.
- �� Option C → Underestimates the field.
Substitution
- Application
- Use the axial field formula and compare with the center field.
- Final Logic
- B=27/125B_0=0.216B_0
3-4-5 Triangle Gives 27 By 125
12 When calculating the magnetic field on the axis of a circular current loop, the components perpendicular to the axis cancel out because:
�� Symmetry is responsible for cancellation. �� Opposite elements produce opposite transverse components. �� Axial components add together.
Consider two diametrically opposite current elements on a circular loop. Each current element produces a magnetic field at a point on the axis. The components perpendicular to the axis are equal in magnitude but opposite in direction. Therefore, these perpendicular components cancel each other. The components along the axis point in the same direction and hence add together. This symmetry is the reason why the resultant magnetic field on the axis of a circular loop is directed entirely along the axis. Hence, the cancellation occurs due to the contribution of diametrically opposite current elements. Therefore, Option B is correct.
- �� Option A → The loop need not be infinitely long.
- �� Option C → Uniformity of current density is not the reason.
- �� Option D → The angle is not always zero.
Logical Analysis
- Application
- Analyze the contributions from diametrically opposite current elements.
- Final Logic
- Opposite transverse components cancel; axial components add.
Opposite Elements Cancel Sideways Parts
13 For a circular loop, the field at the center is B. If the current is scaled by x and the radius by y, the new field is:
�� Field at center is proportional to I/R. �� Scale current and radius separately. �� Apply proportionality directly.
The magnetic field at the center of a circular loop is: B=μ_0I/2R Hence, B∝I/R If current becomes: I^'=xI and radius becomes: R^'=yR then B^'=μ_0(xI)/2(yR)B^'=x/y(μ_0I/2R)B^'=x/yB Therefore, Option A is correct.
- �� Option B → Radius dependence is ignored.
- �� Option C → Reverses the proportionality.
- �� Option D → Addition is not involved.
Concept Application
- Application
- Apply the proportionality B∝I/R.
- Final Logic
- B^'=x/yB
Current Up Radius Down Field Up
14 Incorrect statement regarding a tightly wound coil of N turns.
B=μ_0NI/2R
�� Fields from turns add together. �� More turns produce a stronger field. �� B∝N.
For a tightly wound coil having N turns, the magnetic field at the center is: B=μ_0NI/2R This equation shows that the magnetic field is directly proportional to the number of turns. Therefore, increasing N increases the magnetic field. Since the coil is tightly wound, all turns may be considered to have approximately the same radius. The magnetic field contributions from all turns add according to the principle of superposition. Hence, the statement that the magnetic field decreases as N increases is incorrect. Therefore, Option C is correct.
- �� Option A → Correct consequence of superposition.
- �� Option B → Valid approximation for tightly wound coils.
- �� Option D → Correct NCERT formula.
NCERT Recall
- Application
- Recall the magnetic field expression for an N-turn coil.
- Final Logic
- B∝N
More Turns More Field
15 When evaluating a circular current loop, the upper side and lower side can respectively be thought of as a:
�� A current loop behaves like a magnetic dipole. �� One face acts as north pole. �� The opposite face acts as south pole.
A circular current loop produces a magnetic field pattern similar to that of a bar magnet. Using the right-hand thumb rule, one face of the loop acts as a north pole while the opposite face acts as a south pole. Magnetic field lines emerge from the north face and enter the south face, just as they do for a bar magnet. Therefore, when describing the two sides of a current loop, they can be regarded as magnetic north and south poles. Hence, the upper side and lower side can respectively be considered a north pole and a south pole. Therefore, Option A is correct.
- �� Option B → East pole is not a magnetic pole.
- �� Option C → Positive and negative are electric concepts.
- �� Option D → Axial and radial describe directions, not poles.
Concept Application
- Application
- Compare a circular current loop with a magnetic dipole.
- Final Logic
- Current Loop Behaves Like a Bar Magnet.
Loop Acts Like Magnet
16 Identify the correct statements regarding the right-hand thumb rule for circular loops.
Statements:
1. The thumb points in the direction of the electric field.
2. Fingers point in the direction of the current.
3. The thumb gives the direction of the magnetic field.
4. Curl the palm around the wire.
�� Fingers follow current direction. �� Thumb indicates magnetic field direction. �� Used for circular current loops.
The right-hand thumb rule is used to determine the direction of the magnetic field produced by a current-carrying circular loop. The fingers of the right hand are curled around the loop in the direction of the current. When this is done, the extended thumb points along the direction of the magnetic field through the axis of the loop. The thumb does not indicate the direction of the electric field. Instead, it indicates the magnetic field direction and helps identify the north pole of the loop. Therefore, statements 2, 3 and 4 are correct while statement 1 is incorrect.
- �� Option A → Statement 1 is incorrect.
- �� Option B → Statement 1 is incorrect.
- �� Option C → Statement 4 is also correct.
NCERT Recall
- Application
- Recall the standard right-hand thumb rule for current loops.
- Final Logic
- Fingers → Current; Thumb → Magnetic Field.
Fingers Follow Current Thumb Shows Field
17 Identify the correct statements about the straight segments in a bent wire setup evaluated at the center of the arc.
Statements:
1. dl and r are parallel.
2. Their contribution to the magnetic field is zero.
3. The cross product dl×r=0.
4. They contribute a field equal to μ_0I/4πR.
�� Parallel vectors produce zero cross product. �� sin0^∘=0. �� No magnetic field contribution is produced.
According to the Biot-Savart law: dB=μ_0/4πIdlsinθ/r^2 For the straight segments connected to the arc, the displacement vector r from the current element to the center lies along the same line as dl. Therefore: θ=0^∘ and dl×r=0 Since: sin0^∘=0 the magnetic field contribution from these straight portions is zero. Hence, statements 1, 2 and 3 are correct while statement 4 is incorrect.
- �� Option A → Statement 4 is incorrect.
- �� Option C → Statement 4 is incorrect.
- �� Option D → Includes statement 4, which is incorrect.
Concept Application
- Application
- Apply the Biot-Savart law for θ=0^∘.
- Final Logic
- Parallel vectors ⇒ Zero cross product ⇒ Zero field.
Parallel Means No Magnetic Field
18 Identify the correct statements regarding points where the magnetic field contribution from an element dl is identically zero.
Statements:
1. Along the line of the element where θ=0^∘.
2. Where the displacement vector r is perpendicular to dl.
3. Where sinθ=0.
4. Directly on the straight line extending from the element.
�� Biot-Savart law contains sinθ. �� Zero angle gives zero field. �� Points on the element's line contribute no field.
The Biot-Savart law is: dB=μ_0/4πIdlsinθ/r^2 When the observation point lies on the extension of the current element, θ=0^∘ Therefore, sin0^∘=0 and dB=0 Hence, magnetic field contribution becomes zero. When r is perpendicular to dl, θ=90^∘ and the field becomes maximum rather than zero. Therefore, statements 1, 3 and 4 are correct.
- �� Option A → Statement 2 is incorrect and statement 3 is omitted.
- �� Option B → Statement 2 is incorrect.
- �� Option D → Statement 2 is incorrect.
NCERT Recall
- Application
- Use the sinθdependence of the Biot-Savart law.
- Final Logic
- θ=0^∘⇒sinθ=0⇒dB=0
Zero Angle Zero Field
19 In a semi-circular arc carrying a steady current, the contribution to the magnetic field at the center resembles a full circular loop because:
�� All elemental fields point in the same direction. �� Contributions add directly. �� Symmetry simplifies the calculation.
For every current element on a semi-circular arc, the displacement vector from the element to the center is radial. The current element dl is tangential to the arc and therefore perpendicular to r. As a result, dl×r has the same direction for all current elements. Therefore, every elemental magnetic field contribution points either into or out of the plane and all contributions add constructively. This is why the magnetic field due to a semi-circular arc can be obtained by summing all elemental contributions exactly as in the circular-loop derivation. Hence, Option A is correct.
- �� Option B → A semicircle does not completely surround the center.
- �� Option C → Displacement vectors do not cancel the field.
- �� Option D → Symmetry alone does not explain the addition of fields.
Concept Application
- Application
- Analyze the direction of dl×r for every current element.
- Final Logic
- All elemental fields point in the same direction and add.
Same Direction Means Addition
20 A current element of length Δx=2 cm is at the origin carrying 5 A. The magnetic field at a distance of 1 m on the y-axis is:
μ_0/4π=10^(-7) T m A^(-1)
�� Use the Biot-Savart law. �� Convert centimetres into metres. �� Here θ=90^∘.
The Biot-Savart law gives: dB=μ_0/4πIΔxsinθ/r^2 Given: I=5 AΔx=2 cm=2×10^(-2) mr=1 mθ=90^∘ Substituting: dB=10^(-7)×(5)(2×10^(-2))(1)/1^2dB=10^(-7)×10^(-1)dB=10^(-8) T Therefore, dB=1×10^(-8) T Hence, Option D is correct.
- �� Option A → Five times the correct value.
- �� Option B → Twice the correct value.
- �� Option C → Ignores the factor 2×10^(-2).
Substitution
- Application
- Substitute the values directly into the Biot-Savart law.
- Final Logic
- 10^(-7)×5×2×10^(-2)=10^(-8) T
Convert Centimetres Before Calculation
