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NEET PHYSICSEasy

A wire carrying a current II along the positive xx-axis has length LL. It is kept in a magnetic field B⃗=(2i^+3j^−4k^)\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) T. The magnitude of the magnetic force acting on the wire is:

A

3IL\sqrt{3}IL

B

3IL3IL

C

5IL\sqrt{5}IL

D

5IL5IL

Step-by-Step Solution

  1. Identify Formula: The magnetic force F⃗\vec{F} exerted on a straight wire of length LL carrying current II in a magnetic field B⃗\vec{B} is given by the cross product: F⃗=I(L⃗×B⃗)\vec{F} = I(\vec{L} \times \vec{B})
  2. Define Vectors: The wire lies along the positive xx-axis, so the length vector is L⃗=Li^\vec{L} = L\hat{i}. The magnetic field is given as B⃗=2i^+3j^−4k^\vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k}.
  3. Calculate Cross Product: Substitute the vectors into the formula: F⃗=I[(Li^)×(2i^+3j^−4k^)]\vec{F} = I [ (L\hat{i}) \times (2\hat{i} + 3\hat{j} - 4\hat{k}) ] Distribute the cross product (recall i^×i^=0\hat{i} \times \hat{i} = 0, i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, and i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}): F⃗=I[L(2)(i^×i^)+L(3)(i^×j^)−L(4)(i^×k^)]\vec{F} = I [ L(2)(\hat{i} \times \hat{i}) + L(3)(\hat{i} \times \hat{j}) - L(4)(\hat{i} \times \hat{k}) ] F⃗=I[0+3L(k^)−4L(−j^)]\vec{F} = I [ 0 + 3L(\hat{k}) - 4L(-\hat{j}) ] F⃗=I(3Lk^+4Lj^)=IL(4j^+3k^)\vec{F} = I (3L\hat{k} + 4L\hat{j}) = IL (4\hat{j} + 3\hat{k})
  4. Calculate Magnitude: The magnitude of the force vector ∣F⃗∣|\vec{F}| is: ∣F⃗∣=IL42+32|\vec{F}| = IL \sqrt{4^2 + 3^2} ∣F⃗∣=IL16+9=IL25|\vec{F}| = IL \sqrt{16 + 9} = IL \sqrt{25} ∣F⃗∣=5IL|\vec{F}| = 5IL
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