YouTube expands paid channels pilot to partners with at least 10,000 existing subscribers, adds mobile support

/// YouTube expands paid channels pilot to partners with at least 10,000 existing subscribers, adds mobile support

October 23, 2013  |  Blog

YouTube today announced it is expanding its paid channels pilot to all eligible partners that have at least 10,000 existing subscribers. Furthermore, YouTube viewers can now discover and subscribe to paid channels from YouTube’s mobile website.

All paid channels will continue to have a 14-day free trial, so potential viewers still have the opportunity to explore content on paid channels. If you’re interested in creating a paid channel, visit your Account Features page to enable the feature after you read the next paragraph.

To be eligible to create a YouTube paid channel you must meet the following requirements:

Your account is in good standing.

You meet the general criteria for YouTube partnership.

You have verified your account by phone.

You have an approved AdSense account linked to your YouTube account.

You own a free channel with at least 10,000 active subscribers.

You are located in one of these countries: Australia (AU), Brazil (BR), Canada (CA), France (FR), Japan (JP), Mexico (MX), South Korea (KR), Spain (ES), the United Kingdom (GB), and the United States (US).

Support for Mexico was just added today, and YouTube says an unknown number of additional countries will be launching in the “coming months.” These YouTube channels come with monthly subscription fees starting at $0.99 and are Google’s latest attempt to help content creators monetize their content.

Back in May, YouTube launched paid channels to select partners, letting them create paid channels and allowing users in the aforementioned countries to subscribe to them. The company still won’t say, however, when the new feature will be available for all.

Link: YouTube expands paid channels pilot to partners with at least 10,000 existing subscribers, adds mobile support

The Next Web – Emil  Protalinski


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