📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!

    Question

    A thin conducting rod of length 3 m is aligned

    vertically (along the z-axis) and is moving in space with a velocity vector v = (i + 4j) m/s within a uniform magnetic field given by B = (3i+j) Wb/m2 What is the magnitude of the emf induced between the ends of the rod?
    A 36 V Correct Answer Incorrect Answer
    B 50 V Correct Answer Incorrect Answer
    C 48 V Correct Answer Incorrect Answer
    D 33 V Correct Answer Incorrect Answer

    Solution

    The induced emf in a moving conductor in a magnetic field is given by the formula: ϵ=(v×B) ⋅ L where v is the velocity vector of the conductor, B is the magnetic field vector, and L is the vector representing the length and orientation of the conductor.     Given: ·        Length of the rod, l=3m. ·        The rod is aligned vertically along the z-axis, so the length vector is L=3km. ·        Velocity vector, v=i+4jm/s. ·        Magnetic field vector, B=3i+jWb/m First, calculate the cross-product v×B: Now, calculate the induced emf ϵ=(v×B) ⋅ L: ϵ=(−11k) ⋅ (3k) ϵ=(−11)×(3)(k ⋅ k) Since k ⋅ k=1, ϵ=−33V. The magnitude of the induced emf is ∣ ϵ ∣ = ∣ −33 ∣ =33V.

    Practice Next
    More Physics GS Questions

    Relevant for Exams:

    ask-question