CUET UG Physics Booster Test 2 -The Lorentz Force

The Lorentz force law describes the total electromagnetic force acting on a charged particle: F=q[E+(v×B)] The electric part qE can do work on a charged particle because it may have a component along the direction of motion. In contrast, the magnetic force q(v×B)is always perpendicular to the velocity of the particle. Since work done is given by the scalar product of force and displacement, a force perpendicular to motion performs no work. Consequently, the magnetic field cannot increase or decrease the kinetic energy of the particle. It can only change the direction of motion. The magnetic force component contains a vector cross product and depends on charge, velocity and magnetic field. The complete electromagnetic force law was developed through the work of H.A. Lorentz. Therefore, the statement claiming that the magnetic force transfers energy to the particle is incorrect.

The force exerted by an electric field on a charge is: F=qE This force may act along the direction of displacement and therefore can perform work. As a result, electric fields can increase or decrease the kinetic energy of charged particles. The magnetic force is given by: F=q(v×B) Because this force is always perpendicular to the velocity, it remains perpendicular to the displacement. Therefore, W=F⋅s=0 for magnetic forces. This distinction is fundamental in electromagnetism. Electric fields can accelerate charged particles and transfer energy, while magnetic fields only alter the direction of motion. Hence, the work done by an electric field can be non-zero, whereas the work done by a magnetic field is always zero.

The magnetic force on a positive charge is: F=q(v×B) Since the force is obtained from a vector cross product, it always acts perpendicular to the velocity and magnetic field. Thus, statement 1 is correct. If velocity is along the +x-axis and magnetic field is along the +y-axis: i×j=k Therefore, the force acts along the +z-axis for a positive charge. Hence, statement 2 is correct. The magnitude of magnetic force is: F=qvBsin⁡θ This force is maximum when θ=90^∘, not when velocity and magnetic field are parallel. Therefore, statement 3 is incorrect. Magnetic force does no work and cannot transfer energy to the particle. Thus, statement 4 is also incorrect.

The magnetic force is: F=q(v×B) For a negative charge, the force direction is opposite to the direction obtained from the right-hand rule. Therefore, statement 1 is correct and statement 3 is also correct. The magnitude of magnetic force is: F=qvBsin⁡θ When the velocity is anti-parallel to the magnetic field: θ=180^∘ and sin⁡180^∘=0 Hence, the magnetic force becomes zero. Therefore, statement 4 is correct. Statement 2 is incorrect because when velocity is perpendicular to the magnetic field, the force becomes maximum rather than zero.

The magnetic force acting on a charged particle is: F=qvBsin⁡θ This expression clearly shows that magnetic force is directly proportional to velocity. If a charge is stationary: v=0 Therefore, F=0 regardless of the magnetic field strength or the orientation of the field. This is an important distinction between electric and magnetic fields. An electric field can exert a force on a stationary charge through: F=qE whereas a magnetic field requires motion of the charge to produce a force. The definition of one Tesla is based on a charge of 1 C moving with a speed of 1 m/s perpendicular to the field and experiencing a force of 1 N. A stationary charge does not satisfy this condition because the magnetic force becomes zero. Hence, Option B is correct.

The magnetic force acting on a charged particle moving in a magnetic field is given by: F=qvBsin⁡θ Since the electron moves perpendicular to the magnetic field: θ=90^∘ and sin⁡90^∘=1 Therefore, F=qvB Substituting the values: F=(1.6×10^(-19))(3×10^7)(6×10^(-4))F=28.8×10^(-16)F=2.88×10^(-15) N Unit Verification (C)(m/s)(T)=N Thus, the force experienced by the electron is 2.88×10^(-15) N. Hence, Option A is correct.

The magnetic force is given by: F=q(v×B) Suppose v=v_xi+v_yj and B=Bk Then, v×B=(v_xi+v_yj)×Bk Using vector identities: i×k=-jj×k=i Therefore, the resulting force contains only x and y components. No z-component appears. Since the force is perpendicular to the magnetic field, and the field is along z, the force must remain entirely within the xy-plane. Thus, Option D is correct.

The magnetic force acting on a charged particle is: F=qvBsin⁡θ When the particle moves parallel or anti-parallel to the magnetic field: θ=0^∘ or 180^∘ Since sin⁡0^∘=0 and sin⁡180^∘=0 the magnetic force becomes zero. From vector algebra, parallel vectors produce a zero cross product: v×B=0 Thus, statements 2 and 3 are correct. Statement 1 is incorrect because F=qvB only when θ=90^∘. Statement 4 is incorrect because circular motion requires a non-zero centripetal force, which is absent in this case. Therefore, statements 2 and 3 are correct.

For a proton: v=+iB=+j Therefore, i×j=k The magnetic force acts along the +z-axis. Thus, item 1 matches with b. For an electron, the force direction is opposite because of the negative charge. Hence, item 2 matches with a. A positive charge follows the right-hand rule directly, so item 3 matches with c. A negative charge experiences a force opposite to the right-hand rule prediction, so item 4 matches with d. Therefore, the correct matching is: 1-b, 2-a, 3-c, 4-d Hence, Option C is correct.

The magnetic force is: F=q(v×B) The vector product (v, B)determines the force direction for a positive charge. The screw rule or right-hand rule is used to find this direction. For a negative charge moving under identical conditions, the magnitude of the force remains the same because the magnitude depends on ∣q∣. However, the sign of the charge reverses the direction of the force vector. Therefore, if the force direction predicted by the right-hand rule is considered positive for a positive charge, then the force on the negative charge will be in the opposite direction. Hence, the signs are opposite for positive and negative charges.

The magnetic force acting on a charged particle is given by: F=q(v×B) Since the force is obtained from a vector cross product, it is always perpendicular to the velocity of the particle. Therefore, statement 1 is correct. Work done by a force is given by: W=F⋅s Because magnetic force remains perpendicular to displacement, the dot product becomes zero. Hence, no work is done by the magnetic field, making statement 2 correct. Since no work is done, kinetic energy remains constant. As kinetic energy depends on speed, the magnitude of velocity also remains unchanged. Thus, statement 3 is correct. Statement 4 is incorrect because magnetic fields do not transfer kinetic energy to charged particles. They only alter the direction of motion. Therefore, statements 1, 2 and 3 are correct.

The magnetic force is: F=q(v×B) Since this force is always perpendicular to velocity, it cannot change the speed of the particle. Therefore, the magnitude of velocity remains constant throughout the motion. Momentum is a vector quantity: p=mv Although the magnitude of momentum remains unchanged, its direction continuously changes because the magnetic force acts sideways to the motion. This is the reason charged particles move in circular or helical paths in magnetic fields. Furthermore, because the magnetic force is perpendicular to displacement, it performs no work. Thus, kinetic energy and speed remain constant. Therefore, the statement claiming that the magnitude of velocity changes is incorrect. Hence, Option D is the correct answer.

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts continuously toward the centre of the circular path. Therefore, it behaves as the required centripetal force. The magnetic force magnitude is: F=qvB and the centripetal force is: F_c=mv^2/r Equating them: qvB=mv^2/r This confirms that the magnetic force provides the centripetal force responsible for circular motion. Therefore, statements 1 and 2 are correct. Statement 3 is incorrect because centripetal force is perpendicular to velocity and performs no work. Consequently, speed does not increase. Statement 4 is incorrect because the magnetic force depends directly on velocity through the term qvB. Hence, only statements 1 and 2 are correct.

For a charged particle moving perpendicular to a magnetic field: r=mv/qB Substituting the given values: r=(9×10^(-31))(3×10^7)/(1.6×10^(-19))(6×10^(-4)) Numerator: =27×10^(-24) Denominator: =9.6×10^(-23) Thus, r=27×10^(-24)/9.6×10^(-23)r=0.28125 mr≈0.28 m Converting to centimetres: r=28 cm Unit Verification kg⋅m/s/C⋅T=m Hence, the radius of the circular path is approximately 28 cm.

The period of revolution is related to angular frequency by: T=2π/ω For a charged particle moving in a magnetic field, the angular frequency is: ω=qB/m Substituting: T=2π/qB/m Dividing by a fraction is equivalent to multiplying by its reciprocal: T=2π(m/qB) Therefore, T=2πm/qB This result shows that the period depends only on the mass, charge and magnetic field. It does not depend on the particle's speed or orbit radius. This property is fundamental to cyclotron operation. Hence, Option A is correct.

The cyclotron frequency is given by: ν=qB/2πm This expression contains only the charge of the particle, the magnetic field strength and the mass of the particle. Neither velocity, kinetic energy nor orbit radius appears in the formula. As a charged particle gains energy inside a cyclotron, its circular path radius increases. However, the frequency of revolution remains unchanged. This allows the alternating electric field inside the cyclotron to operate at a fixed frequency while continuously accelerating the particle. Statement 2 is therefore correct because cyclotron frequency is independent of energy. Statement 3 is also correct because this property forms the fundamental operating principle of the cyclotron. Statements 1 and 4 are incorrect because frequency does not depend on initial velocity or orbit radius. Hence, statements 2 and 3 are correct.

The magnetic force acting on a charged particle is: F=q(v×B) The force depends only on the component of velocity perpendicular to the magnetic field. The component parallel to the magnetic field experiences no magnetic force because the angle between the vectors is 0^∘. Therefore, the parallel component continues unchanged. At the same time, the perpendicular component produces circular motion under the action of the magnetic force. The superposition of uniform motion along the magnetic field and circular motion around the magnetic field lines produces a helical trajectory. This type of motion is commonly observed for charged particles moving in magnetic fields in laboratories, plasma devices and cosmic environments. Hence, Option D is correct.

In helical motion, the velocity of the charged particle can be resolved into: • v_⊥→ perpendicular component producing circular motion • v_∥→ parallel component producing forward motion The pitch of the helix is defined as the distance traveled along the magnetic field direction during one complete revolution. Using the basic relation: Distance=Speed×Time the pitch becomes: p=v_∥T The perpendicular component contributes only to the circular part of the motion and does not contribute to advancement along the magnetic field direction. Therefore, the distance moved along the magnetic field in one revolution is equal to v_∥×T. Hence, Option A is correct.

The magnetic force acting on a straight current-carrying conductor placed in a uniform magnetic field is: F=BIlsin⁡θ Since the magnetic field is perpendicular to the current: θ=90^∘ and sin⁡90^∘=1 Thus, F=BIl Substituting the given values: F=(0.5)(2)(1.5)F=1.5 N Unit Verification (T)(A)(m)=N Therefore, the magnetic force acting on the rod is 1.5 N. Hence, Option B is correct.

For a straight conductor placed in a uniform magnetic field, the magnetic force is directly calculated using: F=I(l×B) Therefore, item 1 matches with b. For a conductor of arbitrary shape, the wire is divided into infinitesimal elements dl. The elemental force is: dF=I(dl×B) Thus, item 2 matches with a and item 3 matches with c. The total force on a complex conductor is obtained by vector integration: F=I∫(dl×B) Hence, item 4 matches with d. Therefore, the correct matching is: 1-b, 2-a, 3-c, 4-d Thus, Option D is correct.