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Magnetics question

2017 · 8 Apr · Shift 1 · Q57
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  5. /2017 · 8 Apr · Shift 1 · Q57

Magnetics question

2017 · 8 Apr · Shift 1 · Q57

JEE MainPhysicsMagneticsMCQ+4 / −1
In a certain region static electric and magnetic fields exist. The magnetic field is given by B→=B0(i^+2j^−4k^)\overrightarrow B = {B_0}\left( {\widehat i + 2\widehat j - 4\widehat k} \right)B=B0​(i+2j​−4k). If a test charge moving with a velocity υ→=υ0(3i^−j^+2k^)\overrightarrow \upsilon = {\upsilon _0}\left( {3\widehat i - \widehat j + 2\widehat k} \right)υ=υ0​(3i−j​+2k) experiences no force in that region, then the electric field in the region, in SI units, is :
  1. A
    E→=−υ0 B0(3i^−2j^−4k^)\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {3\widehat i - 2\widehat j - 4\widehat k} \right)E=−υ0​B0​(3i−2j​−4k)
  2. B
    E→=−υ0 B0(i^+j^+7k^)\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {\widehat i + \widehat j + 7\widehat k} \right)E=−υ0​B0​(i+j​+7k)
  3. C
    E→=υ0 B0(14j^+7k^)\overrightarrow E = {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)E=υ0​B0​(14j​+7k)
  4. D
    E→=−υ0 B0(14j^+7k^)\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)E=−υ0​B0​(14j​+7k)
View written solutionFree

Correct answer: D

  1. For a charged particle to experience no force in electric and magnetic fields,
F⃗=q(E⃗+v⃗×B⃗)=0\vec F = q(\vec E + \vec v \times \vec B)=0F=q(E+v×B)=0

Since q≠0q \neq 0q=0 for a test charge,

E⃗=− v⃗×B⃗\vec E = -\,\vec v \times \vec BE=−v×B
  1. Given:
B⃗=B0(i^+2j^−4k^),v⃗=v0(3i^−j^+2k^)\vec B = B_0(\hat i + 2\hat j - 4\hat k), \qquad \vec v = v_0(3\hat i - \hat j + 2\hat k)B=B0​(i^+2j^​−4k^),v=v0​(3i^−j^​+2k^)

So,

v⃗×B⃗=v0B0 (3i^−j^+2k^)×(i^+2j^−4k^)\vec v \times \vec B = v_0B_0\,(3\hat i - \hat j + 2\hat k) \times (\hat i + 2\hat j - 4\hat k)v×B=v0​B0​(3i^−j^​+2k^)×(i^+2j^​−4k^)
  1. Compute the cross product using determinant form:
v⃗×B⃗=v0B0∣i^j^k^3−1212−4∣\vec v \times \vec B = v_0B_0 \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -1 & 2 \\ 1 & 2 & -4 \end{vmatrix}v×B=v0​B0​​i^31​j^​−12​k^2−4​​

Expanding,

v⃗×B⃗=v0B0[i^{(−1)(−4)−2(2)}−j^{3(−4)−2(1)}+k^{3(2)−(−1)(1)}]\vec v \times \vec B = v_0B_0\left[\hat i\{(-1)(-4)-2(2)\} - \hat j\{3(-4)-2(1)\} + \hat k\{3(2)-(-1)(1)\}\right]v×B=v0​B0​[i^{(−1)(−4)−2(2)}−j^​{3(−4)−2(1)}+k^{3(2)−(−1)(1)}] =v0B0[i^(4−4)−j^(−12−2)+k^(6+1)]= v_0B_0\left[\hat i(4-4) - \hat j(-12-2) + \hat k(6+1)\right]=v0​B0​[i^(4−4)−j^​(−12−2)+k^(6+1)] =v0B0[0i^+14j^+7k^]= v_0B_0\left[0\hat i + 14\hat j + 7\hat k\right]=v0​B0​[0i^+14j^​+7k^]

Thus,

v⃗×B⃗=v0B0(14j^+7k^)\vec v \times \vec B = v_0B_0(14\hat j + 7\hat k)v×B=v0​B0​(14j^​+7k^)
  1. Therefore,
E⃗=−v⃗×B⃗=−v0B0(14j^+7k^)\vec E = -\vec v \times \vec B = -v_0B_0(14\hat j + 7\hat k)E=−v×B=−v0​B0​(14j^​+7k^)
  1. Compare with options:
  • A: Incorrect
  • B: Incorrect
  • C: Incorrect sign
  • D: Correct

Hence the electric field is

E⃗=−v0B0(14j^+7k^)\boxed{\vec E = -v_0B_0(14\hat j + 7\hat k)}E=−v0​B0​(14j^​+7k^)​
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