CUET UG Physics Booster Test 3-Current Elements

The total electromagnetic force acting on a charged particle is known as the Lorentz force and is given by: F=q[E+(v×B)] Therefore, statement 1 is correct. The magnetic force depends on the vector cross product (v, B). If velocity is parallel or anti-parallel to the magnetic field, the angle between them becomes 0^∘or 180^∘. In both cases, the cross product is zero, making statement 2 correct. The electric force is: F_E=qE Its direction depends on the electric field and need not be perpendicular to velocity. Therefore, statement 3 is incorrect. The Lorentz force acts on all charged particles, including electrons and protons. The sign of charge only changes the direction of the force. Hence, statement 4 is incorrect.

Electric fields exert a force: F=qE which may have a component parallel to the motion of the particle. Therefore, electric fields can perform work and transfer energy to charged particles. Hence, statements 1 and 3 are correct. Magnetic force is given by: F=q(v×B) Since this force is always perpendicular to velocity, it performs no work on the particle. Therefore, statement 2 is correct. Because no work is done, the kinetic energy and speed of the particle remain unchanged. Only the direction of motion changes. Consequently, statement 4 is incorrect. Thus, statements 1, 2 and 3 correctly describe the differences between electric and magnetic fields.

The magnetic force on a charged particle is: F=qvBsin⁡θ Since the proton moves perpendicular to the magnetic field: θ=90^∘ and sin⁡90^∘=1 Thus, F=qvB Substituting the given values: F=(1.6×10^(-19))(3×10^7)(6×10^(-4))F=2.88×10^(-15) N The given kinetic energy value is irrelevant because magnetic force depends only on charge, velocity and magnetic field. Therefore, the exact magnetic force acting on the proton is: 2.88×10^(-15) N Hence, Option A is correct.

For a particle moving along the +x-axis in a magnetic field along the +y-axis: i×j=k Therefore, the magnetic force on a positive charge is directed along the +z-axis. Thus, the proton is deflected along +z and item 2 matches with b. An electron has negative charge, so its force direction is opposite to the right-hand rule prediction. Hence, the electron is deflected along the -z-axis and item 1 matches with c. Positive charges follow the right-hand rule directly, whereas negative charges experience force in the opposite direction. Therefore: 1-c, 2-b, 3-d, 4-a which corresponds to Option C.

The magnetic force acting on a charged particle is given by: F=qvBsin⁡θ The force depends directly on the velocity of the particle. For a stationary electron: v=0 Therefore, F=0 The same reasoning applies to a stationary proton. Since its velocity is also zero, the magnetic force acting on it is likewise zero. Although electrons and protons have opposite charges, charge sign affects only the direction of the force when motion exists. If the particle is stationary, no magnetic force is produced regardless of charge sign or magnetic field strength. Hence, both the stationary electron and stationary proton experience zero magnetic force. Therefore, Option B is correct.

The magnitude of magnetic force acting on a charged particle is given by: F=qvBsin⁡θ This expression shows that magnetic force is directly proportional to the velocity of the particle. If the particle is stationary relative to the magnetic field: v=0 Substituting into the formula: F=q(0)Bsin⁡θF=0 Thus, regardless of the charge, magnetic field strength or angle, a stationary charged particle experiences no magnetic force. This is one of the fundamental differences between electric and magnetic fields. An electric field can exert a force on a stationary charge, whereas a magnetic field requires motion of the charge. Hence, Option A is correct.

The magnetic force acting on a charged particle is: F=qvBsin⁡θ where θis the angle between velocity and magnetic field. The vector cross product produces a vector perpendicular to both participating vectors. Therefore, statement A is correct. When θ=0^∘or 180^∘, sin⁡θ=0 and the magnetic force becomes zero. Thus, statement B is correct. When θ=90^∘, sin⁡90^∘=1 which gives the maximum possible magnetic force: F=qvB Therefore, the force does not vanish at 90°. Instead, it reaches its maximum value. Hence, statement C is incorrect. The direction of the cross product is determined using the right-hand rule or screw rule, making statement D correct.

The magnetic force is: F=q(v×B) If the particle moves exactly parallel or anti-parallel to the magnetic field, the angle between v and B is either 0^∘or 180^∘. Therefore, v×B=0 which makes statement 1 correct. Since the cross product is zero, the magnetic force also becomes zero: F=0 Thus, statement 2 is correct. Because no magnetic force acts, the magnetic field cannot alter the particle's motion. The velocity remains unchanged, making statement 4 correct. A helical path requires both perpendicular and parallel velocity components. Since the motion here is purely parallel, the particle moves in a straight line rather than a helix. Hence, statement 3 is incorrect.

The magnetic force on a charged particle is given by: F=q(v×B) The right-hand rule determines the direction of (v, B), which corresponds to the force on a positive charge. For a negative charge such as an electron, the charge q is negative. Multiplying by a negative quantity reverses the force direction. Therefore, the force experienced by a negative charge is opposite to the direction predicted directly by the right-hand rule. This is why electrons and protons moving under identical conditions bend in opposite directions in a magnetic field. Hence, Option B is correct.

The magnetic force is determined by the vector cross product: F=q(v×B) A cross product always produces a vector perpendicular to the plane containing the two original vectors. Therefore, statement 1 is correct. Its magnitude is: ∣v×B∣=vBsin⁡θ making statement 2 correct. If v=+i and B=+j then i×j=k Thus, for a positive charge, the force points along the +z-axis. Hence, statement 3 is correct. Statement 4 is incorrect because electric force is given by qE, which does not involve a vector cross product.

The magnetic force acting on a charged particle is: F=q(v×B) Since the magnetic force is always perpendicular to the instantaneous velocity and displacement, the angle between force and displacement is 90^∘. Work done is given by: W=F⋅s Therefore, W=Fscos⁡90^∘=0 Hence, statement 1 is correct. Since no work is done by the magnetic field, the kinetic energy of the particle remains unchanged. The magnetic field only changes the direction of motion and not the speed. Thus, statement 3 is also correct. Statement 2 is incorrect because work done is never maximum when v and B are parallel; in fact, the magnetic force itself becomes zero in that case. Statement 4 is incorrect because magnetic fields do not transfer energy to charged particles.

In a uniform magnetic field, the magnetic force acts perpendicular to the velocity of the charged particle. Since the force is perpendicular to motion, it performs no work: W=0 Therefore, statement 3 is correct. The magnitude of velocity remains constant because kinetic energy does not change. Hence, statement 1 is incorrect. Momentum is a vector quantity: p=mv Even though its magnitude remains constant, its direction continuously changes due to the magnetic force. Thus, statement 2 is correct. Statement 4 is incorrect because the velocity vector is not completely unaffected. Its magnitude remains constant, but its direction changes continuously. Therefore, statements 2 and 3 are correct.

For circular motion, the required centripetal force is: F_c=mv^2/r This force always points toward the centre of the circular path. In a magnetic field, the magnetic force: F_B=qvB acts perpendicular to the velocity and serves as the centripetal force. Therefore, qvB=mv^2/r Statements A and B are correct. Since the magnetic force is always perpendicular to the velocity, it cannot act parallel to the instantaneous path. If it acted parallel to motion, it would change the speed rather than produce circular motion. Statement D is also correct because the circular motion occurs in a plane perpendicular to the magnetic field. Hence, statement C is incorrect.

For a charged particle moving perpendicular to a magnetic field: r=mv/qB Given: mv=1.0 kg m/sq=0.5 CB=2.0 T Substituting: r=1.0/(0.5)(2.0)r=1.0/1.0r=1.0 m Unit Verification kg m/s/C T=m Thus, the radius of the circular path is exactly 1.0 m. Hence, Option D is correct.

The angular frequency of a charged particle moving in a magnetic field is: ω=qB/m The relation between angular frequency and ordinary frequency is: ω=2πν Substituting: 2πν=qB/m Dividing both sides by 2π: ν=qB/2πm This expression is known as the cyclotron frequency. It depends only on charge, magnetic field and mass, and is independent of velocity and radius. Therefore, the correct relationship is: ν=qB/2πm Hence, Option A is correct.

The frequency of revolution of a charged particle in a uniform magnetic field is: ν=qB/2πm This expression contains only three quantities: • Charge q • Magnetic field B • Mass m Notice that velocity and kinetic energy do not appear in the formula. As the particle gains energy inside a cyclotron, its speed and orbit radius increase, but its frequency of revolution remains constant. This remarkable property allows a fixed-frequency alternating electric field to accelerate the particle repeatedly. The particle always arrives at the accelerating gap at the correct phase of the oscillating electric field. Therefore, the cyclotron frequency is independent of the velocity and energy of the particle, making Option C correct.

When the velocity is completely perpendicular to the magnetic field, the magnetic force acts continuously as a centripetal force. Since the speed remains constant and the force is always perpendicular to velocity, the particle moves in a circular path. Thus: 1 → d When the velocity has both perpendicular and parallel components, the perpendicular component produces circular motion while the parallel component causes uniform motion along the field direction. The combination of these motions creates a helical path. Thus: 2 → c The magnetic force serves as the centripetal force: qvB=mv^2/r Therefore: 3 → b Since no magnetic force acts along the field direction, the parallel component remains unchanged. Therefore: 4 → a Hence the correct matching is: 1-d,  2-c,  3-b,  4-a which corresponds to Option A.

The pitch of a helix is the distance traveled by the particle along the magnetic field direction during one complete revolution. Therefore, p=v_∥T The time period of revolution in a magnetic field is: T=2πm/qB Substituting: p=v_∥(2πm/qB) Thus, p=2πmv_∥/qB The pitch increases with the parallel velocity and particle mass, while stronger magnetic fields reduce the pitch. Hence, Option D is correct.

For equilibrium, the upward magnetic force must balance the downward gravitational force. Weight of the wire: W=mgW=(0.2)(9.8)W=1.96 N The magnetic force on the wire is: F=BIL Given: I=2 AL=1.5 m For suspension: BIL=mg Substituting: B(2)(1.5)=1.963B=1.96B=0.653 TB≈0.65 T Therefore, the required magnetic field is 0.65 T.

For a straight conductor placed in a uniform magnetic field, the magnetic force is directly calculated using: F=I(l×B) This expression is sufficient because the length vector and magnetic field remain well-defined throughout the conductor. For a conductor of arbitrary shape, different segments may have different orientations. Therefore, the wire is divided into infinitesimal elements dl. The force on each element is: dF=I(dl×B) The total force is then obtained by integration: F=I∫(dl×B) Hence, a straight rod uses a simple cross-product expression, whereas an arbitrary conductor requires integration of elemental forces. Therefore, Option B is correct.