1. The deflection in a moving coil galvanometer is
Correct Answer: (b) Directly proportional to the number of turns in the coil
The deflection θ in a moving coil galvanometer is given by θ = nABI/k, where n is number of turns, A is area, B is magnetic field, I is current, and k is torsional constant. Thus, deflection is directly proportional to the number of turns.
2. A moving coil sensitive galvanometer gives at once much more deflection. To control its speed of deflection
Correct Answer: (c) A small copper wire should be connected across its terminals
Connecting a small copper wire across the terminals creates eddy currents that oppose the motion of the coil, providing damping to control the speed of deflection.
3. A current carrying rectangular coil is placed in a uniform magnetic field. In which orientation, the coil will not tend to rotate
Correct Answer: (b) The magnetic field is perpendicular to the plane of the coil
When the magnetic field is perpendicular to the plane of the coil, the torque τ = nIABsinθ becomes zero (θ=0°), so the coil won't rotate. In all other orientations, there will be some component of torque.
4. A current loop of magnetic moment \( \vec{m} \) is placed in a uniform magnetic field \( \vec{B} \). The potential energy of the system is minimum when the angle between \( \vec{m} \) and \( \vec{B} \) is:
Correct Answer: (a) 0°
The potential energy U of a magnetic dipole in a magnetic field is given by \( U = -\vec{m} \cdot \vec{B} = -mB\cosθ \). This is minimized when θ=0° (cosθ=1), meaning the magnetic moment is aligned with the field.
5. A coil carrying electric current is placed in uniform magnetic field, then
Correct Answer: (a) Torque is formed
A current-carrying coil in a uniform magnetic field experiences a torque given by \( \vec{τ} = \vec{m} × \vec{B} \) where \( \vec{m} \) is the magnetic moment. An EMF is only induced when there is a change in magnetic flux, which doesn't occur for a stationary coil in a uniform field.
6. A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of 0.1 Wb/m². The amount of work done in rotating it through 180° from its equilibrium position will be
Correct Answer: (a) 0.1 J
Work done \( W = mB(\cosθ_1 - \cosθ_2) \). Here, magnetic moment \( m = nIA = 50 × 2 × π×(0.04)^2 = 0.5 A·m² \). Initial θ₁=0°, final θ₂=180°.
\( W = 0.5 × 0.1 × (1 - (-1)) = 0.1 J \)
\( W = 0.5 × 0.1 × (1 - (-1)) = 0.1 J \)
7. A current carrying wire is bent into a circular loop of radius R. If it is placed in a uniform magnetic field B with its plane perpendicular to the field, the net magnetic force on the loop is:
Correct Answer: (c) Zero
For a closed current loop in a uniform magnetic field, the net magnetic force is always zero, though there may be a torque. This is because the forces on opposite elements of the loop cancel each other out.
8. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field \( \vec{B} \), such that \( \vec{B} \) is perpendicular to the plane of the loop. The magnetic force acting on the loop is
Correct Answer: (c) Zero
In a uniform magnetic field, the net force on any closed current loop is zero, regardless of its shape. This is because the forces on different segments cancel each other out, though there may be a torque if the magnetic moment is not aligned with the field.
9. Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert
Correct Answer: (a) An attractive force on each other
Parallel currents in the same direction attract each other, while currents in opposite directions repel. The force per unit length is given by \( F/L = \frac{μ_0 I_1 I_2}{2πd} \), where d is the separation between wires.
10. Two long and parallel wires are at a distance of 0.1 m and a current of 5 A is flowing in each of these wires. The force per unit length due to these wires will be
Correct Answer: (a) \( 5 × 10^{-5} N/m \)
Force per unit length between parallel currents is \( F/L = \frac{μ_0 I_1 I_2}{2πd} = \frac{(4π×10^{-7}) × 5 × 5}{2π × 0.1} = 5 × 10^{-5} N/m \).
11. 3 A of current is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a magnetic field of strength 500 gauss and makes an angle of 30° with the direction of the field. It experiences a force of magnitude
Correct Answer: (a) \( 3 × 10^{-2} N \)
Magnetic force \( F = I L B \sinθ \). Here, B = 500 gauss = 500 × 10⁻⁴ T = 0.05 T, L = 0.4 m, I = 3 A, θ = 30°.
\( F = 3 × 0.4 × 0.05 × \sin30° = 3 × 0.4 × 0.05 × 0.5 = 3 × 10^{-2} N \)
\( F = 3 × 0.4 × 0.05 × \sin30° = 3 × 0.4 × 0.05 × 0.5 = 3 × 10^{-2} N \)
12. A square loop of side 10 cm carrying a current of 1 A is placed in a magnetic field of 0.5 T such that the plane of the loop makes an angle of 60° with the field direction. The magnitude of the torque acting on the loop is:
Correct Answer: (a) \( 2.5 × 10^{-3} N·m \)
Torque \( τ = nIAB\sinθ \). Here, n=1, I=1 A, A=(0.1 m)²=0.01 m², B=0.5 T, θ=90°-60°=30° (angle between magnetic moment and field).
\( τ = 1 × 1 × 0.01 × 0.5 × \sin30° = 2.5 × 10^{-3} N·m \)
\( τ = 1 × 1 × 0.01 × 0.5 × \sin30° = 2.5 × 10^{-3} N·m \)
13. To make the field radial in a moving coil galvanometer
Correct Answer: (c) Poles are cylindrically cut
In a moving coil galvanometer, the pole pieces are made cylindrical to produce a radial magnetic field. This ensures that the plane of the coil is always parallel to the magnetic field, making the torque maximum and independent of the deflection angle.
14. A current carrying wire of length L is bent into a semicircular arc of radius R. If placed in a uniform magnetic field B perpendicular to its plane, the magnetic force on the wire is:
Correct Answer: (a) \( 2IRB \)
For a current-carrying wire in a magnetic field, the net force is the same as if the wire was straight between its endpoints. Here, the effective length is the diameter (2R), so \( F = I(2R)B = 2IRB \).
15. Through two parallel wires A and B, 10 and 2 ampere of currents are passed respectively in opposite direction. If the wire A is infinitely long and the length of the wire B is 2 m, the force on the conductor B, which is situated at 10 cm distance from A will be
Correct Answer: (a) \( 8 × 10^{-5} N \)
Force between parallel wires is \( F = \frac{μ_0 I_1 I_2 L}{2πd} \). Here, μ₀=4π×10⁻⁷, I₁=10 A, I₂=2 A, L=2 m, d=0.1 m.
\( F = \frac{(4π×10^{-7}) × 10 × 2 × 2}{2π × 0.1} = 8 × 10^{-5} N \) (repulsive since currents are opposite)
\( F = \frac{(4π×10^{-7}) × 10 × 2 × 2}{2π × 0.1} = 8 × 10^{-5} N \) (repulsive since currents are opposite)
16. The unit of electric current "ampere" is the current which when flowing through each of two parallel wires spaced 1 m apart in vacuum and of infinite length will give rise to a force between them equal to
Correct Answer: (a) \( 2 × 10^{-7} N/m \)
The ampere is defined such that two parallel wires 1 m apart carrying 1 A each experience a force of exactly \( 2 × 10^{-7} N \) per meter of length. This comes from the formula \( F/L = \frac{μ_0 I_1 I_2}{2πd} \) with μ₀=4π×10⁻⁷ N/A².
17. A moving coil galvanometer has N number of turns in a coil of effective area A, it carries a current I. The magnetic field B is radial. The torque acting on the coil is
Correct Answer: (d) \( NABI \)
The torque on a moving coil galvanometer is given by \( τ = NABI \), where N is number of turns, A is area, B is magnetic field, and I is current. The radial magnetic field ensures maximum torque (sinθ=1).
18. A small coil of N turns has area A and a current I flows through it. The magnetic dipole moment of this coil will be
Correct Answer: (d) \( NIA \)
The magnetic dipole moment \( \vec{m} \) of a current loop is given by \( \vec{m} = NI\vec{A} \), where N is number of turns, I is current, and \( \vec{A} \) is area vector perpendicular to the plane of the loop.
19. A rectangular loop carrying a current i is placed in a uniform magnetic field B. The area enclosed by the loop is A. If there are n turns in the loop, the torque acting on the loop is given by
Correct Answer: (b) \( niAB \sinθ \)
The torque on a current loop in a magnetic field is \( τ = \vec{m} × \vec{B} = mB\sinθ \), where \( m = niA \) is the magnetic moment. Thus, \( τ = niAB\sinθ \), where θ is the angle between the magnetic moment and the field.
20. A charged particle moves in a uniform magnetic field. The work done by the magnetic force on the particle depends on:
Correct Answer: (d) None of these, as magnetic force does no work
The magnetic force \( \vec{F} = q(\vec{v} × \vec{B}) \) is always perpendicular to the velocity \( \vec{v} \), so the work done \( W = \vec{F} \cdot \vec{d} = 0 \). The magnetic force can change the direction of motion but not the kinetic energy.
21. Two thin long parallel wires separated by a distance b are carrying a current i amp each. The magnitude of the force per unit length exerted by one wire on the other is
Correct Answer: (a) \( \frac{μ_0 i^2}{2πb} \)
The force per unit length between two parallel current-carrying wires is given by \( F/L = \frac{μ_0 I_1 I_2}{2πd} \). Here, \( I_1 = I_2 = i \) and \( d = b \), so \( F/L = \frac{μ_0 i^2}{2πb} \).
22. A current carrying circular loop is freely suspended by a long thread. The plane of the loop will point in the direction
Correct Answer: (b) North-south
A current loop behaves like a magnetic dipole. In Earth's magnetic field, it will align itself such that its magnetic moment points along the north-south direction, just like a compass needle.
23. A proton moving with velocity \( \vec{v} \) enters a uniform magnetic field \( \vec{B} \). The angle between \( \vec{v} \) and \( \vec{B} \) is θ. The path of the proton will be:
Correct Answer: (d) All of the above
The path depends on the angle θ between velocity and magnetic field:
- θ = 0° or 180°: Straight line (no force as \( \vec{F} = q\vec{v} × \vec{B} = 0 \))
- θ = 90°: Circular path (constant perpendicular force)
- 0° < θ < 90°: Helical path (combination of circular and linear motion)
24. A circular coil of diameter 7 cm has 24 turns of wire carrying current of 0.75 A. The magnetic moment of the coil is
Correct Answer: (b) 0.2 A·m²
Magnetic moment \( m = NIA \). Here, N=24, I=0.75 A, radius r=3.5 cm=0.035 m, so area \( A = πr^2 = π×(0.035)^2 ≈ 0.00385 m² \).
\( m = 24 × 0.75 × 0.00385 ≈ 0.07 A·m² \). (Note: The options don't match the calculation - this might be an error in the original question)
\( m = 24 × 0.75 × 0.00385 ≈ 0.07 A·m² \). (Note: The options don't match the calculation - this might be an error in the original question)
25. Two long parallel wires carrying equal current separated by 1 m, exert a force of \( 2 × 10^{-7} N/m \) on one another. The current flowing through them is
Correct Answer: (b) 1.0 A
From \( F/L = \frac{μ_0 I_1 I_2}{2πd} \), with \( I_1 = I_2 = I \), \( d = 1 m \), and \( F/L = 2 × 10^{-7} N/m \):
\( 2 × 10^{-7} = \frac{(4π×10^{-7}) × I × I}{2π × 1} \) ⇒ \( I^2 = 1 \) ⇒ \( I = 1 A \)
\( 2 × 10^{-7} = \frac{(4π×10^{-7}) × I × I}{2π × 1} \) ⇒ \( I^2 = 1 \) ⇒ \( I = 1 A \)
26. The expression for the torque acting on a coil having area of cross-section A, number of turns n, placed in a magnetic field of strength B, making an angle θ with the normal to the plane of the coil, when a current i is flowing in it, will be
Correct Answer: (a) \( niAB \sinθ \)
The torque on a current loop is \( τ = \vec{m} × \vec{B} = mB\sinθ \), where \( m = niA \) is the magnetic moment. Here θ is the angle between the magnetic moment (normal to the plane) and the field, so \( τ = niAB\sinθ \).
27. A charged particle moves in a region of uniform magnetic field along a helical path of radius 5 cm and pitch 20 cm. The ratio of its tangential velocity to its perpendicular velocity is:
Correct Answer: (b) 2:1
In helical motion:
- Radius \( r = \frac{mv_⊥}{qB} \) (depends on perpendicular velocity)
- Pitch \( p = v_∥T = v_∥(\frac{2πm}{qB}) \) (depends on parallel velocity)
28. Two parallel wires of length 9 m each are separated by a distance 0.15 m. If they carry equal currents in the same direction and exerts a total force of 30 × 10⁻⁷ N on each other, then the value of current must be
Correct Answer: (d) 0.5 amp
Total force \( F = \frac{μ_0 I^2 L}{2πd} \). Here, \( F = 30 × 10^{-7} N \), \( L = 9 m \), \( d = 0.15 m \):
\( 30 × 10^{-7} = \frac{(4π×10^{-7}) × I^2 × 9}{2π × 0.15} \) ⇒ \( I^2 = 0.25 \) ⇒ \( I = 0.5 A \)
\( 30 × 10^{-7} = \frac{(4π×10^{-7}) × I^2 × 9}{2π × 0.15} \) ⇒ \( I^2 = 0.25 \) ⇒ \( I = 0.5 A \)
29. A current loop is placed in a non-uniform magnetic field. What can be said about the net force and torque on the loop?
Correct Answer: (d) Both force and torque may exist
In a non-uniform magnetic field:
- The net force on the loop may not be zero because different parts experience different field strengths
- The torque depends on the orientation of the loop relative to the field direction
- Both force and torque can exist simultaneously in a non-uniform field
30. A current of 10 ampere is flowing in a wire of length 1.5 m. A force of 15 N acts on it when it is placed in a uniform magnetic field of 2 tesla. The angle between the magnetic field and the direction of the current is
Correct Answer: (a) 30°
Magnetic force \( F = I L B \sinθ \). Here, \( F = 15 N \), \( I = 10 A \), \( L = 1.5 m \), \( B = 2 T \):
\( 15 = 10 × 1.5 × 2 × \sinθ \) ⇒ \( \sinθ = 0.5 \) ⇒ \( θ = 30° \)
\( 15 = 10 × 1.5 × 2 × \sinθ \) ⇒ \( \sinθ = 0.5 \) ⇒ \( θ = 30° \)
31. Two long parallel copper wires carry currents of 5 A each in opposite directions. If the wires are separated by a distance of 0.5 m, then the force between the two wires is
Correct Answer: (b) \( 10^{-5} N/m \), repulsive
Force per unit length \( F/L = \frac{μ_0 I_1 I_2}{2πd} = \frac{(4π×10^{-7}) × 5 × 5}{2π × 0.5} = 10^{-5} N/m \). Since currents are in opposite directions, the force is repulsive.
32. A charged particle is projected in a region where both electric field \( \vec{E} \) and magnetic field \( \vec{B} \) exist. For the particle to move with constant velocity, the condition is:
Correct Answer: (c) \( \vec{E} + \vec{v} × \vec{B} = 0 \)
For constant velocity (no acceleration), the net force must be zero: \( \vec{F} = q(\vec{E} + \vec{v} × \vec{B}) = 0 \). This requires \( \vec{E} = -\vec{v} × \vec{B} \), meaning the electric force must exactly balance the magnetic force.
33. Two parallel conductors A and B of equal lengths carry currents I and 10 I, respectively, in the same direction. Then
Correct Answer: (b) A and B will attract each other with same force
According to Newton's Third Law, the forces between A and B are equal and opposite. Since currents are in the same direction, the wires attract each other with equal magnitude forces.
34. A straight wire of length 0.5 metre and carrying a current of 1.2 ampere placed in a uniform magnetic field of induction 2 Tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is
Correct Answer: (b) 1.2 N
Magnetic force \( F = I L B \sinθ \). Here, θ=90° (perpendicular), so \( F = 1.2 × 0.5 × 2 × 1 = 1.2 N \).
35. If two streams of protons move parallel to each other in the same direction, then they
Correct Answer: (c) Attract each other
Two parallel currents in the same direction attract each other magnetically. The electrostatic repulsion between protons is typically much weaker than the magnetic attraction at relativistic speeds.
36. Two parallel wires in free space are 10 cm apart and each carries a current of 10 A in the same direction. The force one wire exerts on the other per metre of length is
Correct Answer: (a) \( 2 × 10^{-4} N/m \), attractive
Force per unit length \( F/L = \frac{μ_0 I_1 I_2}{2πd} = \frac{(4π×10^{-7}) × 10 × 10}{2π × 0.1} = 2 × 10^{-4} N/m \). Since currents are in same direction, the force is attractive.
37. A conductor in the form of a right angle ABC with AB = 3 cm and BC = 4 cm carries a current of 10 A. There is a uniform magnetic field of 5 T perpendicular to the plane of the conductor. The force on the conductor will be
Correct Answer: (c) 2.5 N
The effective length is the straight-line distance from A to C: \( \sqrt{3^2 + 4^2} = 5 \) cm = 0.05 m.
Force \( F = I L B = 10 × 0.05 × 5 = 2.5 N \).
Force \( F = I L B = 10 × 0.05 × 5 = 2.5 N \).
38. A current loop is placed in a magnetic field such that its plane makes an angle of 30° with the field direction. If the torque experienced by the loop is τ, what will be the torque when the angle is changed to 60°?
Correct Answer: (b) \( \sqrt{3}τ \)
Torque \( τ = mB\sinθ \). Initially \( τ = mB\sin30° = mB/2 \).
At 60°, \( τ' = mB\sin60° = mB\sqrt{3}/2 = \sqrt{3}τ \).
At 60°, \( τ' = mB\sin60° = mB\sqrt{3}/2 = \sqrt{3}τ \).
39. The coil of a galvanometer consists of 100 turns and effective area of 1 square cm. The restoring couple is \( 10^{-8} N·m/degree \). The magnetic field between the pole pieces is 5 T. The current sensitivity of this galvanometer will be
Correct Answer: (a) \( 5 × 10^4 degree/ampere \)
Current sensitivity \( S_I = \frac{NAB}{k} \), where k is restoring couple per degree.
Here, N=100, A=1 cm²=10⁻⁴ m², B=5 T, k=10⁻⁸ N·m/degree.
\( S_I = \frac{100 × 10^{-4} × 5}{10^{-8}} = 5 × 10^6 \) degree/(N·m) × \( \frac{N·m}{A} \) = \( 5 × 10^4 \) degree/ampere.
Here, N=100, A=1 cm²=10⁻⁴ m², B=5 T, k=10⁻⁸ N·m/degree.
\( S_I = \frac{100 × 10^{-4} × 5}{10^{-8}} = 5 × 10^6 \) degree/(N·m) × \( \frac{N·m}{A} \) = \( 5 × 10^4 \) degree/ampere.
40. A charged particle moves in a circular path of radius r in a uniform magnetic field. If its kinetic energy is doubled while the magnetic field strength is halved, the new radius of its path will be:
Correct Answer: (c) 2r
Radius \( r = \frac{mv}{qB} \). Kinetic energy \( K = \frac{1}{2}mv^2 \), so \( v ∝ \sqrt{K} \).
If K doubles, v increases by \( \sqrt{2} \). If B is halved, \( r ∝ v/B \).
New radius \( r' = \frac{\sqrt{2}v}{B/2} = 2\sqrt{2}r \). (Note: There seems to be inconsistency in options - this requires re-evaluation)
If K doubles, v increases by \( \sqrt{2} \). If B is halved, \( r ∝ v/B \).
New radius \( r' = \frac{\sqrt{2}v}{B/2} = 2\sqrt{2}r \). (Note: There seems to be inconsistency in options - this requires re-evaluation)
41. A current carrying loop is placed in a uniform magnetic field. The torque on the loop is maximum when the angle between the plane of the loop and the magnetic field is:
Correct Answer: (a) 0°
Torque \( τ = mB\sinθ \), where θ is the angle between the magnetic moment (normal to the plane) and the field. Maximum torque occurs when θ=90°, which corresponds to the plane of the loop being parallel to the field (0° between plane and field).
42. A triangular loop of side l carries a current I. It is placed in a magnetic field B such that the plane of the loop is in the direction of B. The torque on the loop is
Correct Answer: (a) Zero
When the plane of the loop is in the direction of B, the magnetic moment \( \vec{m} \) (perpendicular to the plane) is perpendicular to \( \vec{B} \). However, for a triangular loop in this orientation, the net torque is zero because the forces on the sides cancel out.
43. The pole pieces of the magnet used in a pivoted coil galvanometer are
Correct Answer: (d) Cylindrical surfaces of a horse-shoe magnet
In a moving coil galvanometer, the pole pieces are made cylindrical (often part of a horse-shoe magnet) to produce a radial magnetic field. This ensures the plane of the coil is always parallel to the field, making the torque maximum and independent of the deflection angle.
44. The sensitiveness of a moving coil galvanometer can be increased by decreasing
Correct Answer: (d) The couple per unit twist of the suspension
Sensitivity \( S = \frac{NAB}{k} \), where k is the torsional constant (couple per unit twist). To increase sensitivity, we can decrease k (use a weaker suspension) or increase N, A, or B. Of the options given, decreasing k is the correct choice.
45. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same kinetic energy. The ratio of their radii of circular paths is:
Correct Answer: (b) 1:2
Radius \( r = \frac{mv}{qB} \). Kinetic energy \( K = \frac{1}{2}mv^2 \), so \( v = \sqrt{2K/m} \).
For proton: \( m_p = m \), \( q_p = e \) ⇒ \( r_p = \frac{\sqrt{2Km}}{eB} \)
For alpha: \( m_α = 4m \), \( q_α = 2e \) ⇒ \( r_α = \frac{\sqrt{2K·4m}}{2eB} = \frac{\sqrt{2Km}}{eB} \)
Ratio \( r_p : r_α = 1 : 1 \). (Note: There seems to be inconsistency in options - this requires re-evaluation)
For proton: \( m_p = m \), \( q_p = e \) ⇒ \( r_p = \frac{\sqrt{2Km}}{eB} \)
For alpha: \( m_α = 4m \), \( q_α = 2e \) ⇒ \( r_α = \frac{\sqrt{2K·4m}}{2eB} = \frac{\sqrt{2Km}}{eB} \)
Ratio \( r_p : r_α = 1 : 1 \). (Note: There seems to be inconsistency in options - this requires re-evaluation)
46. A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the axis of the loop is
Correct Answer: (a) An end-on position
For a magnetic dipole, the axis corresponds to the "end-on" position, while the plane of the loop corresponds to the "broad side-on" position. Points on the axis of the loop are in the end-on position relative to the equivalent dipole.
47. A small cylindrical soft iron piece is kept in a galvanometer so that
Correct Answer: (a) A radial uniform magnetic field is produced
The soft iron core in a galvanometer serves to strengthen and make the magnetic field radial, ensuring maximum and uniform torque on the coil regardless of its angular position.
48. A straight conductor carries a current of 5 A. An electron travelling with a speed of \( 10^6 m/s \) parallel to the wire at a distance of 0.1 m from the conductor, experiences a force of
Correct Answer: (a) \( 1.6 × 10^{-18} N \)
Magnetic field due to wire \( B = \frac{μ_0 I}{2πr} = \frac{(4π×10^{-7}) × 5}{2π × 0.1} = 10^{-5} T \).
Force on electron \( F = qvB\sinθ = (1.6×10^{-19}) × 10^6 × 10^{-5} × 1 = 1.6 × 10^{-18} N \) (θ=90°).
Force on electron \( F = qvB\sinθ = (1.6×10^{-19}) × 10^6 × 10^{-5} × 1 = 1.6 × 10^{-18} N \) (θ=90°).
49. A circular loop of area \( 0.01 m^2 \) carrying a current of 10 A, is held perpendicular to a magnetic field of intensity 0.1 T. The torque acting on the loop is
Correct Answer: (a) Zero
When the loop is perpendicular to the field, the magnetic moment \( \vec{m} \) is parallel to \( \vec{B} \), so torque \( τ = mB\sinθ = 0 \) (since θ=0°).
50. Magnetic dipole moment of a rectangular loop is
Correct Answer: (d) Perpendicular to plane of loop and proportional to area of loop
The magnetic dipole moment \( \vec{m} \) of a current loop is given by \( \vec{m} = I\vec{A} \), where \( \vec{A} \) is area vector perpendicular to the plane of the loop. Thus, it's proportional to both current and area, and its direction is perpendicular to the plane of the loop.