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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Magnetic Effects of Current

BITSAT 2024 · Physics · Magnetic Effects of Current

The straight wire A B carries a current I . The ends of the wire...

The straight wire ABA B carries a current II. The ends of the wire subtend angles θ1\theta_1 and θ2\theta_2 at the point PP as shown in figure. The magnetic field at the point PP is The straight wire A B carries a current I . The ends of the wire subtend angles θ_1 and θ_2 at the point P as shown in

Options

  1. Aμ0I4πd(sin⁡θ1−sin⁡θ2)\frac{\mu_0 I}{4 \pi d} \left(\sin \theta_1-\sin \theta_2\right)
  2. Bμ0I4πd(sin⁡θ1+sin⁡θ2)\frac{\mu_0 I}{4 \pi d} \left(\sin \theta_1+\sin \theta_2\right)
  3. Cμ0I4πd(cos⁡θ1−cos⁡θ2)\frac{\mu_0 I}{4 \pi d} \left(\cos \theta_1-\cos \theta_2\right)
  4. Dμ0I4πd(cos⁡θ1+cos⁡θ2)\frac{\mu_0 I}{4 \pi d} \left(\cos \theta_1+\cos \theta_2\right)

Correct answer

A. μ0I4πd(sin⁡θ1−sin⁡θ2)\frac{\mu_0 I}{4 \pi d} \left(\sin \theta_1-\sin \theta_2\right)

Step-by-step solution

The magnetic field at a point PP due to a current carrying straight wire, such that the ends of the wires, makes angle α\alpha and β\beta (see figure) is given by B=μ0I4πd(sin⁡α−sin⁡β)B=\frac{\mu_0 I}{4 \pi d}(\sin \alpha-\sin \beta) where, dd is the distance between the point PP and wire. The straight wire A B carries a current I . The ends of the wire subtend angles θ_1 and θ_2 at the point P as shown in In the given problem, α=θ1\alpha=\theta_1 and \beta=\theta_2\therefore \quad B=\frac{\mu_0 I}{4 \pi d} \left(\sin \theta_1-\sin \theta_2\right)
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