Question

When a current carrying conductor is placed in a magnetic field, it experiences
a mechanical force. What should be the angle between the magnetic field and
the length of the conductor so that the force experienced is:
(i) Zero
(ii) Maximum?​

Answer 1

Answer:

• The angle should be 0° for F = 0
• The angle should be 90° for F(max)

Explanation:

$\rule{300}{1.5}$

From the formula we know that,

$\large\bigstar\;\underline{\boxed{\sf F=I(\ell \times B)}}$

Here,

• F Denotes Force.
• I Denotes current.
• $\ell$ Denotes length.
•  B Denotes Magnetic field.

It can be also written as,

$\longmapsto\sf F=I\times \bigg(\ell\;.\; B\sin\theta\bigg)\\\\\\\longmapsto\sf F=I\times B\; \ell\sin \theta\\\\\\\longmapsto \boxed{\sf F=BI \ell \sin\theta}$

Now,

Case-1

Here Force should be zero.

$\longmapsto \boxed{\sf F=BI \ell \sin\theta}$

Here, the sinθ should be zero, so as to force to be zero (F = 0), we know that sinθ has a value of zero when the value of theta is Zero.

$\longmapsto\sf F=BI\ell\times \sin 0^{\circ}\\\\\\\longmapsto\sf F=BI\ell\times 0\\\\\\\longmapsto\boxed{\sf F=0}$

Therefore,

$\large{\underline{\boxed{\red{\sf \theta=0^{\circ}}}}}$

So, For Force to be zero the angle b/w length and Magnetic field should be 0°.

Case-2

Here Force should be Maximum.

Here, the sinθ should have maximum value, so as to force to be zero (F = F_{max}), we know that sinθ has a maximum value of 1 when the value of theta is 90.

$\longmapsto\sf F=BI\ell\times \sin 90^{\circ}\\\\\\\longmapsto\sf F=BI\ell\times 1\\\\\\\longmapsto\boxed{\sf F=BI\ell}$

Therefore,

$\large{\underline{\boxed{\red{\sf \theta=90^{\circ}}}}}$

So, For Force to be maximum the angle b/w length and Magnetic field should be 90°.

$\rule{300}{1.5}$

Answer 2

Answer:

The angle should be 0° for F = 0

The angle should be 90° for F(max)