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motional emf and induced emf in rotating rod questions with solutions
Practice Electromagnetic Induction for JEE Main, JEE Advanced, NEET, and Classes 11–12. Attempt each question, then open it for the full solution.
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- IntegermediumElectromagnetic Induction
A 1 m long metal rod XY completes the circuit shown. The magnetic field is 0.15 T, the circuit resistance is 5 Ω and the rod moves at 4 m s⁻¹. The force required to maintain constant speed is ______ × 10⁻³ N.
JEE Mains10 April 2023Shift 1Morning Shift
Open question & full solution → - MCQmediumElectromagnetic Induction
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is_________m/s.
(g = acceleration due to gravity)
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JEE Mains 202622 January Morning Shift
Open question & full solution → - MCQmediumElectromagnetic Induction
A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s²) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travells a vertical distance of 200 m is _____ mV.
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JEE Mains 202623 January Morning Shift
Open question & full solution → - MCQmediumElectromagnetic Induction
A metallic rod of length 'L' is rotated with an angular speed of ‘ω’ normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :
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JEE Mains24 January 2023Shift 2Evening Shift
Open question & full solution → - IntegereasyElectromagnetic Induction
A metallic cube of side 15 cm moves along +y at 2 m s⁻¹ in a uniform magnetic field 0.5 T along +z. The equilibrium potential difference between the higher- and lower-potential faces is ______ mV.
JEE Mains11 April 2023Shift 2Evening Shift
Open question & full solution → - NumericmediumElectromagnetic Induction
A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of 60° with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ________ mV. (Take g = 10 m/s²)
JEE Mains 202623 January Morning Shift
Open question & full solution → - NumericeasyElectromagnetic Induction
A rod of length 60 cm rotates with a uniform angular velocity 20 rad s⁻¹ about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is ______ V.
JEE Mains 2024 04 Apr Evening Shift
Open question & full solution → - MCQeasyElectromagnetic Induction
A wire of length 1 m moving with velocity 8 m/s at right angles to a magnetic field of 2 T. The magnitude of induced emf, between the ends of wire will be _______ : Image
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JEE Mains25 January 2023Shift 2Evening Shift
Open question & full solution → - MCQeasyElectromagnetic Induction
An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when moves with a velocity of:
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JEE Mains8 April 2023Shift 2Evening Shift
Open question & full solution → - MCQmediumElectromagnetic Induction
A conducting rod of length L rotates in its own plane about one end with constant angular speed ω. A magnetic field perpendicular to the plane of rotation varies with radial distance r as B(r) = B₀ exp(-λr), where λ is a positive constant. Neglect the centripetal force on electrons in the rod. Which expression gives the motional emf between the ends of the rod?
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JEE Mains 2026 5 April Evening Shift
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