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The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)

Short Answer

Expert verified
The magnetic field strength at a distance \(y/2\) from the long straight current-carrying wire is twice the magnetic field strength at distance \(y\). Therefore, the correct answer is option (c): \(B' = 2B\).

Step by step solution

01

Identify the formula for magnetic field strength

The formula for the magnetic field strength (B) near a long straight current-carrying wire at a distance (r) is given by the following relation from the Biot-Savart law: \[B = \frac{\mu_0 I}{2\pi r}\] where \(\mu_0\) is the permeability of free space, \(I\) is the current through the wire, and \(r\) is the distance from the wire to the point where we wish to calculate the magnetic field strength.
02

Calculate the magnetic field strength at distance 'y'

We are asked to find the magnetic field strength at distance \(y\) and \(y/2\). Let's first find the magnetic field strength at distance \(y\), which is given as \(B\). We can derive the following equation from the formula above: \[B = \frac{\mu_0 I}{2\pi y}\]
03

Calculate the magnetic field strength at distance 'y/2'

Now let's calculate the magnetic field strength at distance \(y/2\), which we can denote as \(B'\). We'll substitute the distance as \(y/2\) in the formula for magnetic field strength: \[B' = \frac{\mu_0 I}{2\pi(\frac{y}{2})}\]
04

Find the relationship between B and B'

We need to find the relationship between \(B\) and \(B'\). Divide the equation of \(B'\) by the equation of \(B\): \[\frac{B'}{B} = \frac{\frac{\mu_0 I}{2\pi(\frac{y}{2})}}{\frac{\mu_0 I}{2\pi y}}\] We can see that \(\mu_0 I\) term cancels out in the numerator and denominator: \[\frac{B'}{B} = \frac{1}{\frac{1}{2}}\]
05

Solve the relationship between B and B'

Now, we can simplify the final equation: \[\frac{B'}{B} = 2\] So, we see that the magnetic field strength at distance \(y/2\) is twice the magnetic field strength at distance \(y\), which is option (c): \(B' = 2B\).

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Most popular questions from this chapter

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

Two similar coils are kept mutually perpendicular such that their centers co- inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown. (a) \(1: \sqrt{2}\) (b) \(1: 2\) (c) \(2: 1\) (d) \(\sqrt{3}: 1\)

1: A charged particle moves in a uniform mag. field. The velocity of the particle at some instant makes an acute angle with the mag. field. The path of the particle will be (a) A straight line (b) A circle (c) A helix with uniform pitch (d) A helix with non-uniform pitch

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\( at an angle of \)60^{\circ}\( to the \)\mathrm{X}$ -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of \(r=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $4 \pi \times 10^{-7} \mathrm{sec}$

A conducting rod of length \(\ell\) [cross-section is shown] and mass \(\mathrm{m}\) is moving down on a smooth inclined plane of inclination \(\theta\) with constant speed v. A vertically upward mag. field \(\mathrm{B}^{-}\) exists in upward direction. The magnitude of mag. field \(B^{-}\) is(a) $[(\mathrm{mg} \sin \theta) /(\mathrm{I} \ell)]$ (b) \([(\mathrm{mg} \cos \theta) /(\mathrm{I} \ell)]\) (c) \([(\mathrm{mg} \tan \theta) /(\mathrm{I} \ell)]\) (d) \([(\mathrm{mg}) /(\mathrm{I} \ell \sin \theta)]\)

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