Q180 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
A conducting rod AB is in contact with metal rails AC and BD, which are $0.25$ m apart in a uniform magnetic field of induction $0.4$ T acting perpendicular to the plane of the figure. Ends C and D are connected through a $5\ \Omega$ resistor. If the rod AB moves to right with a velocity of $5$ m/s, then the magnitude and direction of the current through rod AB is
- A$0.1$ A from A to B
- Correct answerB$0.1$ A from B to A
- C$1$ A from A to B
- D$1$ A from B to A
Solution
The motional emf induced in the moving rod is given by:
$E = Bvl$
Substituting the given values ($B = 0.4\text{ T}$, $v = 5\text{ m/s}$, $l = 0.25\text{ m}$):
$E = 0.4 \times 5 \times 0.25 = 0.5\text{ V}$
The magnitude of the induced current is:
$I = \dfrac{E}{R}$
$I = \dfrac{0.5}{5} = 0.1\text{ A}$
The direction of the induced current can be determined using Fleming's Right-Hand Rule or the Lorentz force on positive charges ($\vec{F} = q(\vec{v} \times \vec{B})$).
With velocity $\vec{v}$ towards the right and magnetic field $\vec{B}$ into the plane of the paper, the force on positive charges is directed upwards.
Therefore, the induced current in the rod AB flows from B to A.
Answer: $0.1$ A from B to A
$E = Bvl$
Substituting the given values ($B = 0.4\text{ T}$, $v = 5\text{ m/s}$, $l = 0.25\text{ m}$):
$E = 0.4 \times 5 \times 0.25 = 0.5\text{ V}$
The magnitude of the induced current is:
$I = \dfrac{E}{R}$
$I = \dfrac{0.5}{5} = 0.1\text{ A}$
The direction of the induced current can be determined using Fleming's Right-Hand Rule or the Lorentz force on positive charges ($\vec{F} = q(\vec{v} \times \vec{B})$).
With velocity $\vec{v}$ towards the right and magnetic field $\vec{B}$ into the plane of the paper, the force on positive charges is directed upwards.
Therefore, the induced current in the rod AB flows from B to A.
Answer: $0.1$ A from B to A
Hence, the correct answer is option: 'B'.