Ncert Exercises & Solutions

Consider a wire carrying a steady current \(I,\) placed in a uniform magnetic field \(B\) perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,

(a)	the motion of charges inside the conductor is unaffected by \(B\) since they do not absorb energy.
(b)	some charges inside the wire move to the surface as a result of \(B\).
(c)	if the wire moves under the influence of \(B\), no work is done by the force.
(d)	if the wire moves under the influence of \(B\), no work is done by the magnetic force on the ions, assumed fixed within the wire.

Choose the correct option:
1. (b), (c)
2. (a), (d)
3. (b), (d)
4. (c), (d)

(3)

Hint: The force acting the charges is perpendicular to the motion of charges.

Step 1: Magnetic forces on a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length is given by;

\vec{F} = i (\vec{l} \times \vec{B})

Step 2: The direction of the force is given by Fleming's left-hand rule. F is perpendicular to the direction of magnetic field B. Therefore, the work done by the magnetic force on the charges is zero.

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